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aontion: C. J. CLAY and SONS,
CAMBRIDGE UNIVERSITY PRESS WAREHOUSE,
AVE MARIA LANE.
(ffilagfloia: 263, ARGTLE STREET.
THE ELEMENTS
OP
COOEDINATE GEOMETRY.
THE ELEMENTS
OF
COOEDINATE aEOMETRY
BY
S. L. LONEY, M.A.,
LATE FELLOW OF SIDNEY SUSSEX COLLEGE, CAMBRIDGE,
PROFESSOR AT THE ROYAL HOLLOWAY COLLEGE.
^^^l^^mU. MASS.
MATH, DEPTi
MACMILLAN AND CO.
AND NEW YOEK.
1895
[All Bights reserved.']
CTambrtlJse:
PRINTED BY J. & C. F. CLAY,
AT THE UNIVEBSITY PRESS.
150553
PKEFACE.
"TN the following work I have tried to present the
elements of Coordinate Geometry in a manner
suitable for Beginners and Junior Students. The
present book only deals with Cartesian and Polar
Coordinates. Within these limits I venture to hope
that the book is fairly complete, and that no proposi-
tions of very great importance have been omitted.
The Straight Line and Circle have been treated
more fully than the other portions of the subject,
since it is generally in the elementary conceptions
that beginners find great difficulties.
There are a large number of Examples, over 1100
in all, and they are, in general, of an elementary
character. The examples are especially numerous in
the earlier parts of the book.
vi PREFACE.
I am much indebted to several friends for reading
portions of the proof sheets, but especially to Mr W.
J. Dobbs, M.A. who has kindly read the whole of the
book and made many valuable suggestions.
For any criticisms, suggestions, or corrections, I
shall be grateful.
S. L. LONEY.
EoTAIi HOLLOWAY COLLEGE,
Egham, Surbey.
July 4, 1895.
CONTENTS.
CHAP. PAGE
I. Introduction. Algebraic Kesults ... 1
II. Coordinates. Lengths of Straight Lines and
Areas of Triangles 8
Polar Coordinates 19
III. Locus. Equation to a Locus 24
IV. The Straight Line. Eect angular Coordinates . 31
Straight line through two points .... 39
Angle between two given straight lines . . 42
Conditions that they may be parallel and per-
pendicular . . . . . . .44
Length of a perpendicular . . " . . 51
Bisectors of angles 58
V The Straight Line. Polar Equations and
Oblique Coordinates . . . . 66
Equations involving an arbitrary constant . . 73
Examples of loci 80
VI. Equations representing two or more Straight
Lines 88
Angle between two lines given by one equation 90
Greneral equation of the second degree . . 94
VII. Transformation of Coordinates . . . 109
Invariants 115
Vlii CONTENTS.
CHAP. PAGE
VIII. The Circle 118
Equation to a tangent 126
Pole and polar 137
Equation to a circle in polar coordinates . .145
Equation referred to oblique axes . . . 148
Equations in terms of one variable . . .150
IX. Systems of Circles 160
Orthogonal circles . . , . . . .160
Kadical axis 161
Coaxal circles 166
X. Conic Sections. The Parabola . 174
Equation to a tangent 180
Some properties of the parabola . . . 187
Pole and polar 190
Diameters 195
Equations in terms of one variable . . .198
XI. The Parabola {continued') .... 206
Loci connected with the parabola . . . 206
Three normals passing through a given point . 211
Parabola referred to two tangents as axes . .217
XII. The Ellipse 225
Auxiliary circle and eccentric angle . . .231
Equation to a tangent . . . . . 237
Some properties of the ellipse .... 242
Pole and polar 249
Conjugate diameters ...... 254
Pour normals through any point . . . 265
Examples of loci 266
XIII. The Hyperbola 271
Asymptotes 284
Equation referred to the asymptotes as axes . 296
One variable. Examples 299
CONTENTS. IX
CHAP. PAGE
XIV. Polar Equation to, a Conic .... 306
Polar equation to a tangent, polar, and normal , 313
XV. General Equation. Tracing of Curves . 322
Particular cases of conic sections .... 322
Transformation of equation to centre as origin 326
Equation to asymptotes 329
Tracing a parabola ...... 332
Tracing a central conic . . . . . . 338
Eccentricity and foci of general conic . 342
XVI. General Equation ...... 349
Tangent 349
Conjugate diameters ...... 352
Conies through the intersections of two conies . 356
The equation S=Xuv 358
General equation to the j)air of tangents drawn
from any point ...... 364
The director circle ....... 365
The foci 367
The axes 369
Lengths of straight lines drawn in given directions
to meet the conic 370
Conies passing through four 23oints . . . 378
Conies touching four lines 380
■ The conic LM=B? 382
XVII. Miscellaneous Propositions .... 385
On the four normals from any point to a central
conic 385
Confocal conies ....... 392
Circles of curvature and contact of the third order . 398
Envelopes 407
Answers . i — xiii
ERKATA.
Page 87, Ex. 27, line 4. For "JR" read " S."
„ 235, Ex. 18, line 3. For "odd" read "even."
,, „ ,, ,, line 5. Dele "and Page 37, Ex. 15."
,, 282, Ex. 3. For "transverse" read "conjugate."
CHAPTER I.
INTRODUCTION.
SOME ALGEBRAIC RESULTS.
1. Quadratic Equations. The roots of the quad-
ratic equation
a'3^ + 6x + c =
may easily be shewn to be
- & + •JlP' — 4ac 1 -b- s/b^ — 4:aG
2i. '^"'^ 2^ •
They are therefore real and unequal, equal, or imaginary,
according as the quantity b^—iac is positive, zero, or negative,
i.e. according as b^ = 4:ac.
2. Relations between the roots of any algebraic equation
and the coejicients of the terms of the equation.
If any equation be written so that the coefficient of the
highest term is unity, it is shewn in any treatise on Algebra
that
(1) the sum of the roots is equal to the coefficient of
the second term with its sign changed,
(2) the sum of the products of the roots, taken two
at a time, is equal to the coefficient of the third term,
(3) the sum of their products, taken three at a time,
is equal to the coefficient of the fourth term with its sign
changed,
and so on.
L. e 1
COORDINATE GEOMETRY.
Ex. 1. If a and /3 be the roots of the equation
b c
ax'^ + bx + c = 0, i.e. x^ + - x + ~ = 0,
a a
we have
b -. ^ c
a + p= — and a^ = -
Ex. 2. If a, j8, and 7 be the roots of the cubic equation
ax^ + bx^ + cx + d=0,
i.e. of
we have
x^+-x^ +-x + - = 0,
a a a
a + p + y:
and
^y + ya + a^=:- ,
o-Pl-
3. It can easily be shewn that the solution of the
equations
a^x + h^y + G^z = 0,
and a^ + h^y + c^z = 0,
IS
X
y
^1^2 ~ ^2^1 ^1^2 ~ ^2^1 '^1^2 ~ ^2^1
Determinant Notation.
4. The quantity-
is called a determinant of the
second order and stands for the quantity a-})^ — aj)^, so that
d-yf d^
^1, h
= Ob^^ — 6»2&i .
\%^\
Exs. (1) ;' | = 2x5-4x3 = 10-12=-2;
!4, 5i
3, -4|
(ii)
-7, -6
= - 3 X ( - 6) - { - 7) X ( - 4) = 18 - 28 = - 10.
DETERMINANTS.
5. The quantity
«!,
»2J
«3
^1,
&2J
^^3
Cl,
^2 5
^3
(1)
is called a determinant of the third order and stands for the
quantity
a. X
^2 J ^3
— a.
^2 5 <^3
&1, &.
+ «o
a> *^3i
61,62
(2),
i.e, by Art. 4, for the quantity
«i (^2^3 - ^3^2) -- «^2 (^1^3 - &3C1) + ^3 (^i^^a - ^2^1)*
i.e. % (62C3 — h..G^ + (^2 (63C1 — 61C3) + «3 (61C2 — 62C1).
6. A determinant of the third order is therefore reduced
to three determinants of the second order by the following
rule :
Take in order the quantities which occur in the first row
of the determinant ; multiply each of these in turn by the
determinant which is obtained by erasing the row and
column to which it belongs ; prefix the sign + and — al-
ternately to the products thus obtained and add the
results.
Thus, if in (1) we omit the row and column to which a^
belongs, we have left the determinant ^'
^ i and this is the
coefficient of a-^ in (2).
Similarly, if in (1) we omit the row and column to which
a^ belongs, we have left the determinant ^'
and this
-'D
with the — sign prefixed is the coefficient of a^ in (2).
7. Ex.
The determinant
1,
-4,
-7,
-2, -3
5,-6
8, -9
X
5,-6
8,-9
-(-2)x
-4,
-7,
--W
■3)x
-4,5
-7,8
= {5x(-9)-8x(-6)}+2x{(-4)(-9)-(-7)(-6)}
-3x{(-4)x8-(-7)x5}
= {-45 + 48} +2(36-42} -3 {-32 + 35}
= 3-12-9= -18.
1—2
COORDINATE GEOMETRY.
8. The quantity
(h.1 ^2> %J ^4
61, &2) hi h
^11 ^25 ^3>
j ^1) ^2 5 ^3) ^4
is called a determinant of the fourth order and stands for
the quantity
«i X
K h, ^4
^2» ^3 J
<^2> ^3> ^4
i^lJ ^35 h
— Clo X \ C-,
^3}
+ 6^3 X
1 1 5 3 3 4
&i, 62J ^4!
C^ cCj_ X
1 ? 2 5 4
&1,
<^2, h
Cl,
^2) Cg
c?i,
»2J <^3
and its value may be obtained by finding the value of each
of these four determinants by the rule of Art. 6.
The rule for finding the value of a determinant of the
fourth order in terms of determinants of the third order is
clearly the same as that for one of the third order given in
Art. 6.
Similarly for determinants of higher orders.
9. A determinant of the second order has two terms.
One of the third order has 3x2, i.e. 6, terms. One of the
fourth order has 4 x 3 x 2, -i.e. 24, terms, and so on.
(1)
(4)
(6)
10. Exs.
2, -3
4, 8
Prove that
= 28. (2)
9, 8, 7j
6, 5, 4 =0.
3, 2, l|
a, h, g
-6,
-4.
7
-9
= 85
!.. (3)
5,
-2,
9,
-3, 7
4,-8
3, -10
-a, b, c
(5)
a, -b, c
=:4a6c.
a, I
, -c
-98.
9, f, c
= abc + 2fgh - ap - bg^ - ch\
ELIMINATION. 5
Elimination.
11. Suppose we have the two equations
aj^x + a^y = (1),
\x +b^y ^0 (2),
between the two unknown quantities x and y. There must
be some relation holding between the four coefficients 6*i, ctaj
bi, and 63 • ^or, from (1), we have
y~ %'
and, from (2), we have - = — =-^ .
y K
X
Equating these two values of - we have
i.e. a-J)^ — ajb^ = (3).
The result (3) is the condition that both the equations
(1) and (2) should be true for the same values of x and y.
The process of finding this condition is called the elimi-
nating of X and y from the equations (1) and (2), and the
result (3) is often called the eliminant of (1) and (2).
Using the notation of Art. 4, the result (3) may be
1 ) '^
This result is obtained from (1) and (2) by taking the
coefficients of x and y in the order in which they occur in
the equations, placing them in this order to form a determi-
nant, and equating it to zero.
written in the form
0.
12. Suppose, again, that we have the three equations
a-^x + a^y + a^^ = (1),
\x+ h^y^ h^z = (2),
and G^x + G^y + C3S = (3),
between the three unknown quantities x, y, and z.
6
COORDINATE GEOMETRY.
By dividing each equation by z we have three equations
X
y
between the two unknown quantities — and -
z z
Two of
%,
^2,
%
&1,
\y
h
Ci,
^2 1
Cs
these will be sufficient to determine these quantities. By
substituting their values in the third equation we shall
obtain a relation between the nine coefficients.
Or we may proceed thus. From the equations (2) and
(3) we have
X __ y _ ^
Substituting these values in (1), we have
«1 (^2^3 - ^3^2) + «2 (^3^1 - ^1^3) + «3 (^1^2 - ^2^1) = 0. . .(4).
This is the result of eliminating cc, 3/, and % from the
equations (1), (2), and (3).
But, by Art. 5, equation (4) may be written in the form
= 0.
This eliminant may be written down as in the last
article, viz. by taking the coefficients of x, y, and z in the
order in which they occur in the equations (1), (2), and (3),
placing them to form a determinant, and equating it to
zero.
13. Ex. What is the value of a so that the equations
ax + 2y + 3z = 0, 2x-3y + 4:Z = 0,
and 5x + 7y-8z=:0
may be simultaneously true ?
Eliminating x, y, and z, we have
a, 2, 3,
2, -3, 41 = 0,
5, 7, -8!
^.e. « [( - 3) ( - 8) - 4 X 7] - 2 [2 X { - 8) - 4 X 5] + 3 [2 X 7 - 5 X ( - 3)]=0,
i.e. «[-4]-2[-36] + 3[29] = 0,
^, ^ 72 + 87 159
so that a= — -, = — ;- .
4 4
ELIMINATION.
14. If again we have the four equations
a-^x + dil/ + cf'zZ + a^u = 0,
h^x + h^y + b^z + b^u = 0,
Ci«; + c^i/ + G^z + c^u = 0,
and djX + d^y + d.^z + d^ — 0,
it could be shewn that the result of eliminating the four
quantities cc, y, z^ and u is the determinant
«1J
^2,
%,
«4
\.
^2,
bz,
^>4
Ci,
^2?
^it
C4
c?i,
C?2,
C?3,
c?.
A similar theorem could be shewn to be true for n
equations of the first degree, such as the above, between
n unknown quantities.
It will be noted that the right-hand member of each of
the above equations is zero.
CHAPTER II.
COORDINATES. LENGTHS OF STRAIGHT LINES AND
AREAS OF TRIANGLES.
15. Coordinates. Let OX and 07 be two fixed
straight lines in the plane of the paper. The line OX is
called the axis of cc, the line OY the axis of y, whilst the
two together are called the axes of coordinates.
The point is called the origin of coordinates or, more
shortly, the origin.
From any point F in the
plane draw a straight line
parallel to OF to meet OX
in M.
The distance OM is called
the Abscissa, and the distance
MP the Ordinate of the point
P, whilst the abscissa and the
ordinate together are called
its Coordinates.
Distances measured parallel to OX are called a?, with
or without a suffix, {e.g.Xj, x.-^... x\ x",...), and distances
measured parallel to OY are called y, with or without a
suffix, (e.g. 2/i, 2/2,--- 2/'. y",---)-
If the distances OM and MP be respectively x and ?/,
the coordinates of P are, for brevity, denoted by the symbol
{x, y).
Conversely, when we are given that the coordinates of
a point P are (x, y) we know its position. For from we
have only to measure a distance OM {—x) along OX and
COORDINATES. 9
then from 21 measure a distance MP {=y) parallel to OY
and we arrive at the position of the point P. For example
in the figure, if OM be equal to the unit of length and
MP= WM, then P is the point (1, 2).
16. Produce XO backwards to form the line OX' and
YO backwards to become OY'. In Analytical Geometry
we have the same rule as to signs that the student has
already met with in Trigonometry.
Lines measured parallel to OX are positive whilst those
measured parallel to OX' are negative ; lines measured
parallel to OY are positive and those parallel to OY' are
negative.
If P2 b® i^ *li® quadrant YOX' and P^M^, drawn
parallel to the axis of y, meet OX' in M^^ and if the
numerical values of the quantities OM^ and J/aPg be a
and h, the coordinates of P are {-a and h) and the position
of Pg is given by the symbol (—a, h).
Similarly, if P3 be in the third quadrant X'OY', both of
its coordinates are negative, and, if the numerical lengths
of Oi/3 and J/3P3 be c and d, then P3 is denoted by the
symbol (— c, — d).
Finally, if P4 lie in the fourth quadrant its abscissa is
positive and its ordinate is negative.
17. Ex. Lay down on "paper the position of the points
(i) (2, -1), (ii) (-3, 2), and (iii) (-2, -3).
To get the first point we measure a distance 2 along OX and then
a distance 1 parallel to OF'; we thus arrive at the required point.
To get the second point, we measure a distance 3 along OX', and
then 2 parallel to OY.
To get the third point, we measure 2 along OX' and then
3 parallel to OT.
These three points are respectively the points P4 , P., , and Pg in
the figure of Art. 15.
18. When the axes of coordinates are as in the figure
of Art. 15, not at right angles, they are said to be Oblique
Axes, and the angle between their two positive directions
OX and 07, i.e. the angle XOY, is generally denoted by
the Greek letter w.
10 COORDINATE GEOMETRY.
In general, it is however found to be more convenient to
take the axes OX and OZat right angles. They are then
said to be Rectangular Axes.
It may always be assumed throughout this book that
the axes are rectangular unless it is otherwise stated.
19. The system of coordinates spoken of in the last
few articles is known as the Cartesian System of Coordi-
nates. It is so called because this system was first intro-
duced by the philosopher Des Cartes. There are other
systems of coordinates in use, but the Cartesian system is
by far the most important.
20. To find the distance between two points whose co-
ordinates are given.
Let Pi and P^ be the two
given points, and let their co-
ordinates be respectively {x^ , y^)
and (a^sj 2/2)-
Draw Pji/i and P^M^ pa-
rallel to OY, to meet OX in
J/j and M^. Draw P^R parallel
to OX to meet M-^P^ in R. q ' M jvT
Then
P^R = M^Mt^ = OM^ - OMc^ = oi^-X2,
RP, = M,P,-M,P, = y,~y,,
and z P^i^Pi = z6>ifiPa-l 80° -PiJfiX^l 80° -<o.
We therefore have [Trigonometry, Art. 164]
P^P^^ = P^R^ + RP^^ - 2P^R . PPi cos P^RP^
- (^1 - x^Y + (2/1 - 2/2)' - 2 (a^i - x^) (2/1 - 2/2) cos (180° - (o)
= (Xi-X2)2 + (yj_y2)2+2(Xi-X2)(yi-y2)COSCO...(l).
If the axes be, as is generally the case, at right angles,
we have <o == 90° and hence cos to = 0.
The formula (1) then becomes
P^P^ - (x, - x^Y + (2/1 - y^Y^
DISTANCE BETWEEN TWO POINTS.
11
SO that in rectangular coordinates the distance between the
two points (x^j y^ and (a-g, 2/2) is
V(Xi - x^)^ + (Yi - y^)^ (2).
Cor. The distance of the point (x^, y-^ from the origin
is Jx^ + 2/1^, the axes being rectangular. This follows from
(2) by making both x^ and y^ equal to zero.
21. The formula of the previous article has been proved for the
case when the coordinates of both the points are all positive.!
Due regard being had to the signs of the coordinates, the formula
will be found to be true for all
points.
As a numerical example, let
Pj be the point (5, 6) and Pg
be the point (-7, -4), so that
we have
and y2= -^.
Then
P^ = 31^0 + OM^ = 7 + 5
and
RPt^ = EM-^ + l/iPj = 4 + 6
= -2/2 + 2/1.
The rest of the proof is as in the last article.
Similarly any other case could be considered.
22. To find tJie coordinates of the point which divides
in a given ratio (ni^ : m^ the line joining two given jyoints
(a?!, 2/1) and (x^, y^).
Yi
O M, M
M,
X
Let Pi be the point {x^, y^), Po the point (x^, y^), and P
the required point, so that we have
12 COORDINATE GEOMETRY.
Let P be the point (sc, y) so that if P^M^, PM, and
P^M^ be drawn parallel to the axis of y to meet the axis of
£C in i/i, Mj and M^, we have
Oi/i = £Ci, M^P^ = y^, OM=x, MP = y, QM^^x^,
and i/^z^a = 2/2-
Draw PiEi and P-Sg, parallel to OX, to meet J/P and
M^P^ in Pi and Pg respectively.
Then PjPi = M^^M^^ OM- OM^ = x-x^,
PR^ = MM^ = OJ/2 - 0M= x,^ - X,
R,P^MP-M,P, = y-y,,
and P2P2 = M^P^ - MP = y^-y.
. From the similar triangles PiPjP and PR^P^ we have
m^ PjP PiRi X — Xt^
m^ PP^ PR^ x^ — x'
, ifv-tt/Uey *T" i/VoOO-i
t.e. x = ^^ .
Again
mi P,P R,P y-y.
- * m^ PP2 P2P2 2/2-2/'
so that mi (3/2 - 3/) = 7^2 {y - 3/1),
and hence y = -^^ ^-^ .
Wi + 7?22
The coordinates of the point which divides PiP^ in-
ternally in the given ratio rrii : tyi^ are therefore
nil + ^2 mi + nig *
If the point Q divide the line P1P2 externally in the
same ratio, i.e. so that P^Q : QP^ :: mj : m^i its coordinates
would be found to be
nil "" ^^2 '^i "■ ^^2
The proof of this statement is similar to that of the
preceding article and is left as an exercise for the student.
LINES DIVIDED IN A GIVEN RATIO. 13
Cor. The coordinates of the middle point of the line
joining {x^, y^ to {x^, y^ are
23. Ex. 1. In any triangle ABC 'prove that
AB^ + AC^ = 2 {AD^ + DG^),
lohere D is the middle point of BG.
Take B as origin, 5C as the axis of x, and a line through B i>er-
pendicular to BC as the axis of y.
Let BG=a, so that G is the point (a, 0), and let A be the point
Then D is the point (|> C> j .
Hence ^D2=ra;i -^Y + i/i^ and DG^=f~y.
Hence 2 (^D^ + DC^) ::= 2 ["x^^ + y^^ - ax^ + ^~|
= 2xi2 + 2yi2_2o.x.^ + a2.
Also ^C'2.= (a;i-a)2 + ?j^2^
and AB^=^x^-\-y^.
Therefore AB'^ + ^(72 = 'Ix^ + 2?/i2 _ 2aa;i + a^.
Hence ^52 + ^(72^2(^2)2 + 2)(72)_
This is the well-known theorem of Ptolemy.
Ex. 2. ABG is a triangle and D, E, and F are the middle points
of the sides BG, GA, and AB ; prove that the point lohich divides AD
internally in the ratio 2 : 1 also divides the lines BE and GF in
the same ratio.
Hence prove that the medians of a triangle meet in a point.
Let the coordinates of the vertices A, J5, and G be (x-^, y-^), (arg, 2/2),
and (ajg, y^) respectively.
The coordinates of D are therefore - ^ ^ and - ^^ .
Let G be the point that divides internally AD in the ratio 2 : 1,
and let its coordinates be x and y.
By the last article
2Xiy "T Xq
^_ 2 ; ^ ^1 + 3^2 + ^3
2+1 3
So ^^2/rt^±^3.
3
14 COORDINATE GEOMETRY.
In the same manner we could shew that these are th^ coordinates
of the points that divide BE and CF in the ratio 2 : 1.
Since the point whose coordinates are
x-^ + x^ + x^ and ^L±^2+l_3
3 3
lies on each of the lines AD, BE, and CF, it follows that these three
lines meet in a point.
This point is called the Centroid of the triangle.
EXAMPLES. I.
Find the distances between the following pairs of points.
1. (2, 3) and (5, 7). 2. (4, -7) and (-1, 5).
3_ ( _ 3, _ 2) and ( - 6, 7), the axes being inclined at 60°.
4. (a, o) and (o, 6). 5. {b + c, c + a) and {c + a, a + b).
6. {a cos a, a sin a) and {a cos |S, a sin /3).
7. {am^^, 2ami) and (am^^ 2am^.
8. Lay down in a figure the positions of the points (1, - 3) and
( - 2,1), and prove that the distance between them is 5.
9. Find the value of x^ if the distance between the points [x^^, 2)
and (3, 4) be 8.
10. A line is of length 10 and one end is at the point (2, - 3) ;
if the abscissa of the other end be 10, prove that its ordinate must be
3 or - 9.
11. Prove that the points (2a, 4a), (2a, 6a), and (2a + s/3a, oa)
are the vertices of an equilateral triangle whose side is 2a.
12. Prove that the points (-2, -1), (1, 0), (4, 3), and (1, 2) are
at the vertices of a parallelogram.
13. Prove that the points (2, -2), (8, 4), (5, 7), and (-1, 1) are
at the angular points of a rectangle.
14. Prove that the point ( - xV. f I) is the centre of the circle
circumscribing the triangle whose angular points are (1, 1), (2, 3),
and ( - 2, 2).
Find the coordinates of the point which
15. divides the line joining the points (1, 3) and (2, 7) in the
ratio 3 : 4.
16. divides the same line in the ratio 3 : - 4.
17. divides, internally and externally, the line joining ( - 1, 2)
to (4, - 5) in the ratio 2 : 3.
[EXS. I.] EXAMPLES. 15
18. divides, internally and externally, the line joining ( - 3, - 4)
to ( - 8, 7) in the ratio 7 : 5.
19. The line joining the points (1, - 2) and ( - 3, 4) is trisected ;
find the coordinates of the points of trisection.
20. The line joining the points ( - 6, 8) and (8, - 6) is divided
into four equal parts ; find the coordinates of the points of section.
21. rind the coordinates of the points which divide, internally
and externally, the line joining the point {a + b, a-h) to the point
(a-&, a + 6) in the ratio a : h.
22. The coordinates of the vertices of a triangle are [x-^, 2/i)»
{x^, y^ and (xg, y^. The line joining the first two is divided in the
ratio I : h, and the line joining this point of division to the opposite
angular point is then divided in the ratio m : Jfc + Z. Find the
coordinates of the latter point of section.
23. Prove that the coordinates, x and y, of the middle point of
the line joining the point (2,3) to the point (3, 4) satisfy the equation
x-y + l=:0.
24. If G be the centroid of a triangle ABC and O be any other
point, prove that
^{GA^-^GB'^+GCr-) = BG^+GA^ + AB\
and OA^ + OB^-\-OG'^=GA^ + GB'^+GG^- + ^Ga\
25. Prove that the lines joining the middle points of opposite
sides of a quadrilateral and the Une joining the middle points of its
diagonals meet in a point and bisect one another.
26. -4, B, G, D... are n points in a plane whose coordinates are
(^i» 2/i)> (^2' 2/2)' (^3> 2/3)j---* -^-S is bisected in the point G-^; G^G is
divided at G^ in the ratio 1:2; G^D is divided at G^ in the ratio
1:3; GgE at G^ in the ratio 1 : 4, and so on until all the points are
exhausted. Shew that the coordinates of the final point so obtained are
^1 + ^2 + 3^3+ •••+^n ^j^^ yi + y^ + Vz+'-'-^Vn
n n
[This point is called the Centre of Mean Position of the n given
points.]
27. Prove that a point can be found which is at the same
distance from each of the four points
(am,, ^) , (a»„ ^) , {am,, ^J , and {am^m^, ^^) .
24. To prove that the area of a trapeziitm, i. e. a quad-
rilateral having two sides parallel, is one half the sum of the
two parallel sides multiplied by the perpendicular distance
between them.
16
COOEDINATE GEOMETRY.
B
Let ABGD be the trapezium having the sides AD and
BC parallel.
Join AC and draw AL perpen-
dicular to BG and ON perpendicular
to AD^ produced if necessary.
Since the area of a triangle is one
half the product of any side and the
perpendicular drawn from the opposite angle, we have
area ABGD = ^ABG ■¥ ^AGD
= l.BG ,AL + l,AD.GN
=^i{BG + AD) X AL.
25. To find the area of the triangle^ the coordinates of
whose angular 'points are given^ the axes being rectangular.
Let ABG be the triangle
and let the coordinates of its
angular points A, B and G be
{x^, 2/1), (a?2, 2/2), and {x^, y^).
Draw AL, BM, and Ci\^ per-
pendicular to the axis of x, and
let A denote the required area.
Then
A == trapezium A LNG + trapezium GNMB — trapezium A LMB
= \LN {LA + NG) + \NM {NG + MB) - \LM (LA + MB),
by the last article,
= i [(^3 - ^1) (2/1 + ys) + {002 - ^z) (2/2 + 2/3) - (^2 - a^i) {Vi + 2/2)]-
On simplifying we easily have
^ ^ I (^172 - XaYi + y^zYz - ^372 + XgYi - x^yg),
or the equivalent form
^ = J [^1 (2/2 - Vz) + ^2 (2/3 - 2/1) + ^3 (2/1 - 2/2)].
If we use the determinant notation this may be written
(as in Art. 5)
^1) 2/1) ^
^2> 2/2? ■'-
, ^3j 2/35 ^
Cor. The area of the triangle whose vertices are the
origin (0, 0) and the points {x^, y-^, {x^, 2/2) is J (^12/2 — ^'22/1)*
AREA OF A QUADRILATERAL,
17
isin w {x^y^
I.e.
fsm wx
26. In the preceding article, if the axes be oblique, the perpen-
diculars AL, BM, and CN, are not equal to the ordinates y^ , y^ , and
2/3, but are equal respectively to yi sin w, 2/2 sin w, and y^ sin w.
The area of the triangle in this case becomes
xu yi, 1
27. In order that the expression for the area in Art. 25 may be
a positive quantity (as all areas necessarily are) the points A, B, and
G must be taken in the order in which they would be met by a
person starting from A and walking round the triangle in such a
manner that the area of the triangle is always on his left hand.
Otherwise the expressions of Art. 25 would be found to be negative.
28. To find the area of a quadrilateral the coordinates
of whose angular foints are given.
Let the angular points of the quadrilateral, taken in
order, be A^ B, C, and D, and let their coordinates be
respectively {x^, y^\ (x.^,, y^), (x^, y^\ and {x^, y^).
Draw ALy BM, CJV, and DH perpendicular to the axis
of X.
Then the area of the quadrilateral
= trapezium ALRD + trapezium BRNO + trapezium GNMB
— trapezium ALMB
= ILR {LA + RD) + IRN (RD + NG) + ^NM {JVC + MB)
-lLM{LA-\-MB)
{(^■4 - ^1) (2/1 + 2/4) + (^3 - ^4) (2/3 + 2/4) + (^2 - ^^) (2/3 + 2/2)
- (a?2 - x^) (2/1 + 2/2)}
{(^12/2 - «^22/i) + fez^s - ^zvi) + {xsy4. - ^42/3) + (^42/i - ^y^)}'
L. 2
_ 1
_ 1
18 COORDINATE GEOMETRY.
29. The above formula may also be obtained by
drawing the lines OA, OB, OC and OD. For the quadri-
lateral ABCn
= AOBG+ AOCD- aOBA- AOAD.
But the coordinates of the vertices of the triangle OBG
are (0, 0), (ajg, 2/2) ^^^ (^35 2/3) ^ hence, by Art. 25, its
area is ^ i^^y-^ — ^zV^)-
So for the other triangles.
The required area therefore
^ h [(«^22/3 - a^32/2) + {^zVa, - ^m) - (^Wi - ^12/2) - {^i2/4 - ^'42/l)l
= i [{^^2 - ^22/1) + (^22/3 - ^32/2) + {^3y4 - ^m) + (^42/l - «^l2/4)]-
In a similar manner it may be shewn that the area
of a polygon of n sides the coordinates of whose angular
points, taken in order, are
(^IJ 2/1/5 V^2) 2/2/3 (^35 2/3)? •••(.'^)i3 2/ra/
is i [(a?i2/2 - a?22/i) + (^22/3 - ^32/2) + • • • + (^n2/i - a'l^/ri)]-
EXAMPLES. II.
Find the areas of the triangles the coordinates of whose angular
points are respectively
1. (1, 3), ( - 7, 6) and (5, - 1). 2. (0, 4), (3, 6) and ( - 8, - 2).
3. (5,2), (-9, -3) and (-3, -5).
4. {a, l> + c), (a, h-c) and {-a, c).
5. {a,c + a), {a, c) and {-a, c-a).
6. {a cos (pi, b sin <p-^), {a cos ^^, b sin ^g) and (a cos ^3, b sin ^g).
7. (a7?ij2^ 2a7?i;^), [arn^^ lam^ and [am^, 2avi^.
8. {awiimg, a(??ii + ??i2)}, {aWaWg, a (7?i2 + %)} ^-^cl
9. lam-,. — I , \am^, — y and -Jawo, — [ .
Prove (by shewing that the area of the triangle formed by them is
zero) that the following sets of three points are in a straight line :
10. (1,4), (3, -2), and (-3,16).
11. (-i, 3), (-5,6), and (-8,8).
12. («. b + c), (6, c + a), and (c, a + b).
[EXS. II.] POLAR COORDINATES. 19
Find the areas of the quadrilaterals the coordinates of whose
angular points, taken in order, are
13. (1,1), (3,4), (5, -2), and (4, -7).
14. (-1, 6), (-3, -9), (5, -8), and (3, 9).
15. If be the origin, and if the coordinates of any two points
Pj and Pg ^6 respectively (%, y^ and [x^, y.^, prove that
OP^ . OP2 . cos P1OP2 = x-^Xc^ + y-^y^ •
30. Polar Coordinates. There is another method,
which is often used, for determining the position of a point
in a plane.
Suppose to be a fixed point, called the origin or
pole, and OX a fixed line, called the initial line.
Take any other point P in the plane of the paper and
join OP. The position of P is clearly known when the
angle XOP and the length OP are given.
[For giving the angle XOP shews the direction in which OP is
drawn, and giving the distance OP tells the distance of P along this
direction.]
The angle XOP which would be traced out by the line
OP in revolving from the initial line OX is called the
vectorial angle of P and the length OP is called its radius
vector. The two taken together are called the polar co-
ordinates of P.
If the vectorial angle be and the radius vector be r, the
position of P is denoted by the symbol (r, 0).
The radius vector is positive if it be measured from the
origin along the line bounding the vectorial angle; if
measured in the opposite direction it is negative.
31. Ex. Construct the positions of the 'points (i) (2, 80°),
(ii) (3, 150°), (iii) (-2, 45°), (iv) ■
(-3, 330°), (v) (3, -210°) and (vi) ff\ /^
(-3, -30°). ^\^ ,->^^
(i) To construct the first point, ^^^^^^^^^"^^
let the radius vector revolve from '/^^^ yC
OX through an angle of 30°, and >/ ""-.,
then mark off along it a distance y ''-•.,
equal to two units of length. We 'p M"
thus obtain the point P^. ^
(ii) For the second point, the radius vector revolves from OX
through 150° and is then in the position OP^ ; measuring a distance 3
along it we arrive at Pg .
2—2
20 COORDINATE GEOMETKY.
(iii) For the third point, let the radius vector revolve from OX
through 45° into the position OL. We have now to measure along
OL a distance - 2, i.e. we have to measure a distance 2 not along OL
but in the opposite direction. Producing iO to Pg, so that OP3 is
2 units of length, we have the required point P3.
(iv) To get the fourth point, we let the radius vector rotate from
OX through 330° into the position OM and measure on it a distance
-3, i.e. 3 in the direction MO produced. We thus have the point P^y
which is the same as the point given by (ii).
(v) If the radius vector rotate through - 210°, it will be in the
position OP2, and the point required is Pg.
(vi)^ For the sixth point, the radius vector, after rotating through
- 30°, is in the position OM: We then measure - 3 along it, i.e. 3 in
the direction MO produced, and once more arrive at the point Pg.
32. It will be observed that in the previous example
the same point P^ is denoted by each of the four sets of
polar coordinates
(3, 150°), (-3, 330°), (3, -210°) and (-3, -30°).
In general it v^ill be found that the same point is given
by each of the polar coordinates
(r, 0), (- r, 180° + 6), {r, - (360° - 6)] and {- r, - (180° - 6%
or, expressing the angles in radians, by each of the co-
ordinates
(r, e\ {-r,7r + 6), {r, - (27r - 0)} and {- r, - (tt - $)}.
It is also clear that adding 360° (or any multiple of
360°) to the vectorial angle does not alter the final position
of the revolving line, so that {r, 6) is always the same point
as (r, ^ + ?i . 360°), where n is an integer.
So, adding 180° or any odd multiple of 180° to the
vectorial angle and changing the sign of the radius vector
gives the same point as before. Thus the point
[-r, ^ + (2n + 1)180°]
is the same point as [— r, 6 + 180°], i.e. is the point [r, 6\
33. To find the length of the straight line joining two
points whose polar coordinates are given.
Let A and B be the two points and let their polar
coordinates be (r^, 6y) and (r^, 6^ respectively, so that
OA^r^, OB = r^, lXOA^O^, and lX0B = 6^,
POLAR COORDINATES.
21
Then (Trigonometry, Art. 164)
AB" - OA'' + OB'' -20 A. OB cos AOB
= r-^ + r^ - 2r-^r^ cos {0^ - 6^.
34. To find the area of a triangle the coordinates of
whose angular points are given.
Let ABC be the triangle and let (r-^, 0^), (r^, 62), and
(rg, ^3) be the polar coordinates of
its angular points.
We have
AABO=AOBC+aOCA
-AOBA (1).
Now
A0BC = i0B,0C sin BOC
[Trigonometry, Art. 198]
= ^r^r^ sin (^3 - $^).
' So A OCA = \0G . OA sin CO A = ^r^r, sin (6, - 6,),
and AOAB^^OA. OB sin AOB = ^r^r^ sin {6^ - 6.^
= - Jn^2 sin (^2 - ^1).
Hence (1) gives
A ABC = J \r<^r^ sin (^3 - 6^ + r^r^ (sin 0-^ - 0^)
+ r^r^ sin {Oo - 0^)].
35. To change from Cartesian Coordinates to Polar
Coordinates, and conversely.
Let P be any point whose Cartesian coordinates, referred
to rectangular axes, are x and y,
and whose polar coordinates, re-
ferred to as pole and OX as
initial line, are (r, 6).
Draw Pit/'perpendicular to OX
so that we have
OM=x, MP = y, LMOP = e,
and OP = r.
From the triangle MOP we
have
x = OM=OPcosMOP = rcosO (1),
y = MP= OPsinMOP ^r smO (2),
r=OP= sJOM^ + MP^^ s/x" + y' (3),
X' O:
22 COORDINATE GEOMETRY,
and
Equations (1) and (2) express the Cartesian coordinates
in terms of the polar coordinates.
Equations (3) and (4) express the polar in terms of the
Cartesian coordinates.
The same relations will be found to hold if P be in any-
other of the quadrants into which the plane is divided by
XOX' and YOT.
Ex. Change to Cartesian coordinates the equations
{!) r = asind, and (2)r=a^cos-.
a
(1) Multiplying the equation by r, it becomes r^rrar sin Q,
i.e. by equations (2) and (3), x^-\-y^ = ay.
(2) Squaring the equation (2), it becomes
r=acos2- = -- (1 + cos^),
i. e. 2^2 = ar + ar cos 6,
i.e. 2{x^ + y^) = a sjx^ + y^-\- ax,
i.e. {2x^ + 2y^-ax)^ = a^{x^ + y^).
EXAMPLES. III.
Lay down the positions of the points whose polar coordinates are
L (3,45°). 2. (-2, -60°). 3. (4,135°). 4. (2,330°).
5. (-1, -180°). 6. (1, -210°). 7. (5, -675°). 8. («> |) •
9. (2a.-^). 10.{-a,l). n.(-2a.4').
Find the lengths of the straight lines joining the pairs of points
whose polar coordinates are
12. (2, 30°) and (4, 120°). 13. (-3, 45°) and (7, 105°).
14. ( «> I ) and
(«"'!) •
[EXS. III.] EXAMPLES. 23
15. Prove that the points (0, 0), (3, |) , and ( 3, ^ J form an equi-
lateral triangle.
Find the areas of the triangles the coordinates of whose angular
points are
16. (1, 30°), (2, 60°), and (3, 90°).
17. (_3, -30°), (5, 150°), and (7, 210°).
18. (-^. i)' («'i)'^^*l(-2«. -y)-
Find the polar coordinates (drawing the figure in each case) of the
points
19. x = J3, y = l. 20. x=-^B, y = l. 21. x=-l, y = l.
Find the Cartesian coordinates (drawing a figure in each case) of
the points whose polar coordinates are
22. (5, I). 23. (-6, I). 24.(5,-^).
Change to polar coordinates the equations
25. x^+y^=a\ 26. y = xtana. 27. x^ + y^ = 2ax.
28. x^-y^=2ay. 29. x^=y^{2a-x). 30. {x^ + y^)^ = a^{x^-y^).
Transform to Cartesian coordinates the equations
31. r=a. 32. ^ = tan-i??i. 33. r = acos^.
34. r = asin2^. 35. ?-2 = a2cos2^. 36. o'^ sin 2d =2a^.
e
2
40. 'T (cos 3^ + sin 3^) = 5h sin 6 cos d
37. r'^G0s2d = a^. 38. r^cos- = a^, 39^ r^=ai sin-.
2i a
CHAPTER III.
LOCUS. EQUATION TO A LOCUS.
36. When a point moves so as always to satisfy a
given condition, or conditions, the path it traces out is
called its Locus under these conditions.
For example, suppose to be a given point in the plane
of the paper and that a point F is to move on the paper so
that its distance from shall be constant and equal to a.
It is clear that all the positions of the moving point must
lie on the circumference of a circle whose centre is and
whose radius is a. The circumference of this circle is
therefore the " Locus" of P when it moves subject to the
condition that its distance from shall be equal to the
constant distance a.
37. Again, suppose A and B to be two fixed points in
the plane of the paper and that a point P is to move in
the plane of the paper so that its distances from A and B
are to be always equal. If we bisect AB in G and through
it draw a straight line (of infinite length in both directions)
perpendicular to AB, then any point on this straight line
is at equal distances from A and B. Also there is no
point, whose distances from A and B are the same, which
does not lie on this straight line. This straight line is
therefore the "Locus" of P subject to the assumed con-
dition.
38. Again, suppose A and B to be two fixed points
and that the point P is to move in the plane of the paper
so that the angle APB is always a right angle. If we
describe a circle on AB as diameter then P may be any
EQUATION TO A LOCUS.
25
point on the circumference of this circle, since the angle
in a semi-circle is a right angle; also it could easily be
shewn that APB is not a right angle except when P lies
on this circumference. The "Locus" of P under the
assumed condition is therefore a circle on -4^ as diameter.
39. One single equation between two unknown quan-
tities x and y, e.g.
x + y = l (1),
cannot completely determine the values of x and y.
*\P6
\Q
vPs
^1?
\1
M
OM
Vs
\ P
Such an equation has an infinite number of solutions.
Amongst them are the followins: :
a: = 0,1
a; =1,1
2/ = 0/'
x= 2,1
y=-l/'
x= 3,1
y=-2/'-
x = —\,\ X — — 2
2/= 2/' y= 3
Let us mark down on paper a number of points whose
coordinates (as defined in the last chapter) satisfy equation
Let OX and OF be the axes of coordinates.
If we mark off a distance OPi (—1) along OY^ we have
a point Pi whose coordinates (0, 1) clearly satisfy equation
If we mark off a distance OP^ (=1) along OX, we have
a point Pg whose coordinates (1, 0) satisfy (1).
26 COORDINATE GEOMETRY.
Similarly the point Pg, (2, — 1), and P4, (3, — 2), satisfy
the equation (1).
Again, the coordinates (—1, 2) of Pg and the coordinates
(—2, 3) of Pg satisfy equation (1).
On making the measurements carefully we should find
that all the points we obtain lie on the line P^P^ (produced
both ways).
Again, if we took any point Q, lying on P^P^, and draw
a perpendicular QM to OX, we should find on measurement
that the sum of its x and y (each taken with its proper
sign) would be equal to unity, so that the coordinates of Q
would satisfy (1).
Also we should find no point, whose coordinates satisfy
(1), which does not lie on P1P2.
All the points, lying on the straight line P1P2, and no
others are therefore such that their coordinates satisfy the
equation (1).
This result is expressed in the language of Analytical
Geometry by saying that (1) is the Equation to the Straight
Line P^P^.
40. Consider again the equation
£c2 + 2/2 = 4 (1).
Amongst an infinite number of solutions of this equa-
tion are the following :
x = 2A x= J?>\ x = J\ x^l \
V3| x = J'I\ x=^i \
2/=2r 2/=x/3r 2/=V2 r y=i r
x = -'2,\ x^-.^?>,\ x = -J2A x = -l, \
2/^0 r 2/^--i /' 3/=-v2r y=-si^)'
53 = 0, \ x=l, \ x=J2, } x=J3)
2/ = -2j ' y
L and
V3/' 2/ = -./2/'-'"% = -l/-
EQUATION TO A LOCUS.
27
1 •
,5i
■i|f
-,..
M ;f^ X
All these points are respectively represented by the
points P^, P^^ F^y ... P^Qj and they
will all be found to lie on the
dotted circle whose centre is
and radius is 2.
Also, if we take any other
point Q on this circle and its
ordinate QM, it follows, since
0M^- + MQ^^0Q' = 4:, that the x
and y of the point Q satisfies (1).
The dotted circle therefore
passes through all the points whose
coordinates satisfy (1).
In the language of Analytical Geometry the equation
(1) is therefore the equation to the above circle.
41. As another example let us trace the locus of the
point whose coordinates satisfy the equation
y'=4^'^ (1);
If we give x a negative value we see that y is im-
possible; for the square of a
real quantity cannot be nega-
tive.
We see therefore that there
are no points lying to the left
of OF.
If we give x any positive
value we see that y has two
real corresponding values which
are equal and of opposite signs.
The following values,
amongst an infinite number of
others, satisfy (1), viz.
x = 0,] x=l, "I x = 2,
y = 0}' y = + 2or-2}' y = 2J2ov -2J2
x=4: I cc=16, I aj = + oo, )
y = + 4: or —4:) ^ '" y — Sor-Sj' '" y = + (X)Ov—ao)'
The origin is the first of these points and P^ and Qi,
Pg and Q^, P^and Q^, ... represent the next pairs of points.
h
28 COORDINATE GEOMETRY.
If we took a large number of values of x and the
corresponding values of ?/, the points thus obtained would
be found all to lie on the curve in the figure.
Both of its branches would be found to stretch away to
infinity towards the right of the figure.
Also, if we took any point on this curve and measured
with sufficient accuracy its x and y the values thus obtained
would be found to satisfy equation (1).
Also we should not be able to find any point, not lying
on the curve, whose coordinates would satisfy (1).
In the language of Analytical Geometry the equation
(1) is the equation to the above curve. This curve is called
a Parabola and will be fully discussed in Chapter X.
42. If a point move so as to satisfy any given condition
it will describe some definite curve, or locus, and there can
always be found an equation between the x and y of any
point on the path.
This equation is called the equation to the locus or
curve. Hence
Def. Equation to a curve. The equation to a
curve is the relation which exists between the coordinates of
any foint on the curve^ and which holds for no other points
except those lying on the curve.
43. Conversely to every equation between x and y it
will be found that there is, in general, a definite geometrical
locus.
Thus in Art. 39 the equation is x + y=\, and the
definite path, or locus, is the straight line P^P^ (produced
indefinitely both ways).
In Art. 40 the equation is x'^ + y'^^ 4, and the definite
path, or locus, is the dotted circle.
Again the equation 2/ = 1 states that the moving point
is such that its ordinate is always unity, i.e. that it is
always at a distance 1 from the axis of x. The definite
path, or locus, is therefore a straight line parallel to OX
and at a distance unity from it.
EQUATION TO A LOCUS. 29
44. In the next chapter it will be found that if the
equation be of the first degree {i.e. if it contain no
products, squares, or higher powers of x and y) the locus
corresponding is always a straight line.
If the equation be of the second or higher degree, the
corresponding locus is, in general, a curved line.
45. We append a few simple examples of the forma-
tion of the equation to a locus.
Ex. 1. A point moves so that the algebraic sum of its distances
from tioo given perpendicular axes is equal to a constant quantity a;
find the equation to its locus.
Take the two straight lines as the axes of coordinates. Let {x, y)
be any point satisfying the given condition. We then ha,wex + y = a.
This being the relation connecting the coordinates of any point
on the locus is the equation to the locus.
It will be found in the next chapter that this equation represents
a straight line.
Ex. 2. The sum of the squares of the distances of a moving point
from the tioo fixed points {a, 0) and {-a, 0) is equal to a constant
quantity 2c^. Find the equation to its locus.
Let (a;, y) be any position of the moving point. Then, by Art. 20,
the condition of the question gives
{ [x - af + /} + { (a; + af + if] = 2c\
i.Ci x^ + y'^ = c^- a~.
This being the relation between the coordinates of any, and every,
point that satisfies the given condition is, by Art. 42, the equation to
the required locus.
This equation tells us that the square of the distance of the point
{x, y) from the origin is constant and equal to c^ - a^, and therefore
the locus of the point is a circle whose centre is the origin.
Ex. 3. A point moves so that its distance from the point (-1,0)
is always three times its distance from the point (0, 2).
Let {x, y) be any point which satisfies the given condition. We
then have
J{x + iy' + {y-0)^=Bj{x - 0)2+ {y - 2)2,
so that, on squaring,
x'^ + 2x + l + y'^=9{x'^ + y'^-4:y + 4),
i.e. 8(a;2 + y2)_2a;-36?/ + 35 = 0.
This being the relation between the coordinates of each, and
every, point that satisfies the given relation is, by Art. 42, the
required equation.
It will be found, in a later chapter, that this equation represents
a circle.
30 COOEDINATE GEOMETRY.
EXAMPLES. IV.
By taking a number of solutions, as in Arts. 39 — 41, sketch
the loci of the following equations :
1. 2x + dy = l0. 2. ^x-y = l. 3. x'^-2ax-Vy'^ = Q.
4. a;2-4aa; + ?/2 + 3a2 = 0. 5. y'^ = x. 6. ^x = y^-^.
7 ^' + ^'=1.
'■4^9
A and B being the fixed points (a, 0) and ( - a, 0) respectively,
obtain the equations giving the locus of P, when
8. PA"^ - P52 _ a constant quantity = 2fc2.
9. PA = nPB, n being constant.
10. P^+PjB = c, a constant quantity.
11. PB^ + PC^=2PA^, C being the point (c, 0).
12. Find the locus of a point whose distance from the point (1, 2)
is equal to its distance from the axis of y.
Find the equation to the locus of a point which is always equi-
distant from the points whose coordinates are
13. (1, 0) and (0, -2). 14. (2, 3) and (4, 5).
15. {a + b, a-h) and {a-b, a + b).
Find the equation to the locus of a point which moves so that
16. its distance from the axis of x is three times its distance from
the axis of y.
17. its distance from the point (a, 0) is always four times its dis-
tance from the axis of y.
18. the sum of the squares of its distances from the axes is equal
to 3.
19. the square of its distance from the point (0, 2) is equal to 4.
20. its distance from the point (3, 0) is three times its distance
from (0, 2).
21. its distance from the axis of x is always one half its distance
from the origin.
22. A fixed point is at a perpendicular distance a from a fixed
straight line and a point moves so that its distance from the fixed
point is always equal to its distance from the fixed line. Find the
equation to its locus, the axes of coordinates being drawn through
the fixed point and being parallel and perpendicular to the given
line.
23. In the previous question if the first distance be (1), always half,
and (2), always twice, the second distance, find the equations to the
respective loci.
CHAPTER IV.
THE STRAIGHT LINE. RECTANGULAR COORDINATES.
46. To find the equation to a straight line which is
parallel to one of the coordinate axes.
Let CL be any line parallel to the axis of y and passing
through a point C on the axis of x such that OG = c.
Let F be any point on this line whose coordinates are
X and y.
Then the abscissa of the point F is
always c, so that
x = c (1).
This being true for every point on
the line CL (produced indefinitely both
ways), and for no other point, is, by
Art. 42, the equation to the line.
It will be noted that the equation does not contain the
coordinate y.
Similarly the equation to a straight line parallel to the
axis oi X is y — d.
Cor. The equation to the axis of a? is 2/ = 0.
The equation to the axis oi y is x — 0.
47. To find the equation to a st7'aight line which cuts
off a given intercept on the axis of y and is inclined at a
given angle to the axis of x.
Let the given intercept be c and let the given angle be a.
X
32 COORDINATE GEOMETRY.
Let C be a point on the axis of y such that OC is c.
Through C draw a straight
line Z(7Z' inclined at an angle
a (= tan~^ m) to the axis of x^
so that tan a — m.
The straight line LCL' is ^^
therefore the straight line ^^
required, and we have to -'l O MX
find the relation between the
coordinates of any point P lying on it.
Draw PM perpendicular to OX to meet in ^ a line
through G parallel to OX.
Let the coordinates of P be cu and ?/» so that OM=x
and MP = y.
Then MP = NP + MN =C]Srt^iia + 00 = m.x + c,
i.e. y = mx+c.
This relation being true for any point on the given
straight line is, by Art. 42, the equation to the straight
line.
[In this, and other similar cases, it could be shewn,
conversely, that the equation is only true for points lying
on the given straight line.]
Cor. The equation to any straight line passing through
the origin, i.e. which cuts off a zero intercept from the axis
of 2/, is found by putting c — O and hence is 3/ = mx.
48. The angle a which is used in the previous article is the
angle through which a straight line, originally parallel to OZ, would
have to turn in order to coincide with the given direction, the rotation
being always in the positive direction. Also m is always the tangent
of this angle. In the case of such a straight line as AB, in the figure
of Art. 50, m is equal to the tangent of the angle PAX (not of the
angle PAO). In this case therefore wi, being the tangent of an obtuse
angle, is a negative quantity.
The student should verify the truth of the equation of the last
article for all points on the straight line LCL', and also for straight
Hnes in other positions, e.g. for such a straight line as A^B^ in the
figure of Art. 59. In this latter case both m and c are negative
quantities.
A careful consideration of all the possible cases of a few proposi-
tions will soon satisfy him that this verification is not always
necessary, but that it is sufficient to consider the standard figure.
THE STRAIGHT LINE.
33
49. Ex. The equation to the straight line cutting off an
intercept 3 from the negative direction of the axis of y, and inclined
at 120° to the axis of a;, is
?/ = a;tanl20° + (-3),
i.e. y= -x^S-S,
i.e. y + x^S + S = 0.
50. 1^0 find the equation to the straight line which cuts
off given i7itercepts a and h from the axes.
Let A and B be on OX and OY respectively, and be
such that OA = a and OB = h.
Join AB and produce it in-
definitely both ways. Let P be
any point (x, y) on this straight
line, and draw PM perpendicular
to OX.
We require the relation that
always holds between x and 3/, so
long as P lies on AB.
By Euc. YI. 4, we have
OM_PB MP _AP
OA~AB' ^"""^ 'OB~AB
OM MP PB + AP
+
OA OB
AB
= 1,
^.e.
X y ^
a D
This is therefore the required equation ; for it is the
relation that holds between the coordinates of any point
lying on the given straight line.
51. The equation in the preceding article may he also obtained
by expressing the fact that the sum of the areas of the triangles OP A
and OPB is equal to OAB, so that
\axy + \hy.x = \ax'b,
and hence
a
52. Ex. 1. Find the equation to the straight line passing
through the -point (3, - 4) and cutting off intercepts, equal but of
opposite signs, from the tioo axes.
Let the intercepts cut off from the two axes be of lengths a and
— a.
34
COORDINATE GEOMETRY.
The equation to the straight line is then
a -a
i.e. x-y = a (1).
Since, in addition, the straight line is to go through the point
(3, -4), these coordinates must satisfy (1), so that
3-(-4) = a,
and therefore a = l.
The required equation is therefore
x-y = 7.
Ex. 2. Find the equation to the straight line lohich passes through
the point (-5, 4) and is such that the portion of it between the axes is
divided by the point in the ratio ofl : 2.
Let the required straight line be - + t = 1. This meets the axes
a b
in the points whose coordinates are {a, 0) and (0, &).
The coordinates of the point dividing the line joining these
points in the ratio 1 : 2, are (Art. 22)
2.a+1.0 , 2.0 + 1.&
If this be the point ( - 5, 4) we have
. 2(1 , b
i.e,-^ and -.
„ 2a , , b
-5:=- and 4=-,
so that a=--Y- and b = 12.
The required straight line is therefore
X y
-i^^l2 '
I.e.
oy
8a; = 60.
53. To find the equation to a straight line in tenns of
the perpendicular let fall upon it from the origin and the
angle that this perpendicular makes with the axis of x.
Let OR be the perpendicular from and let its length
be jo.
Let a be the angle that OR makes
with OX.
Let P be any point, whose co-
ordinates are x and y, lying on AB ',
draw the ordinate PM, and also ML
perpendicular to OR and PN perpen-
dicular to ML.
THE STRAIGHT LINE. 85
Then OL = OMco^a (1),
and LR = NP = MF&inNMP.
But lNMP^W - lNMO= iMOL^a.
LR = MP&m.a (2).
Hence, adding (1) and (2), we have
Oil/ cos a + ifPsin a=OL + LR=OR =79,
i.e. X COS a + y sin a = p.
This is the required equation.
54. In Arts. 47 — 53 we have found that the correspond-
ing equations are only of the first degree in x and y. We
shall now prove that
Any equation of the first degree i7i x and y always repre-
sents a straight line.
For the most general form of such an equation is
Ax + By^C = ^ (1),
where A^ B, and C are constants, i.e. quantities which do
not contain x and y and which remain the same for all
points on the locus.
Let (cCi, 2/1), (a?2) 2/2)) ^iicl (rt's, 2/3) be any three points on
the locus of the equation (1).
Since the point {x-^, y^) lies on the locus, its coordinates
when substituted for x and y in (1) must satisfy it.
Hence Ax^ + Ry^+C-=0 (2).
So Ax^ + Ry^ + C^O (3),
and Axs + £ys+C = (4).
Since these three equations hold between the three quanti-
ties A, B, and C, we can, as in Art. 12, eliminate them.
The result is
^35 2/35 -•-
But, by Art. 25, the relation (5) states that the area of the
triangle whose vertices are (x^, y^), (x^, 3/2)5 ^^^ (^3> 2/3) is
zero.
Also these are any three points on the locus.
3—2
= (5).
36 COORDINATE GEOMETRY.
The locus must therefore be a straight line ; for a curved
line could not be such that the triangle obtained by joining
any three points on it should be zero.
55. The proposition of the preceding article may also be deduced
from Art. 47. For the equation
^a; + % + (7=0
A C
may be written y=- — x-^,
and this is the same as the straight line
y = mx + c,
A ^ C
if ?3i=-— and c = - — .
x> is
But in Art. 47 it was shewn that y = mx + c was the equation to
a straight line cutting off an intercept c from the axis of y and
inclined at an angle tan~^m to the axis of x.
The equation Ax + By + C=0
C
therefore represents a straight line cutting off an intercept - — from
x>
the axis of y and inclined at an angle tan~^ ( - — | to the axis of x.
56. We can reduce the general equation of the first
degree Ax + By + C = (1)
to the form of Art. 53.
For, if p be the perpendicular from the origin on (1)
and a the angle it makes with the axis, the equation to the
straight line must be
X cos a 4- 2/ sin a - /» = (2).
This equation must therefore be the same as (1).
cos a sin a —p
Hence
ABC
p cos a sin a \/cos^ a + sin^ a 1
C -A -B Ja^ + B' sJa^' + B^
Hence
-A . -B ^ C
cos a - - , sm a = , , and p =
s/A^ + B^' \fA-' + B'' sfA^ + B^
The equation (1) may therefore be reduced to the form (2)
by dividing it by JA^ + B^ and arranging it so that the
constant term is negative.
THE STRAIGHT LINE. 37
57. Ex. Reduce to tlie perpendicular form the equation
^ + 2/\/3 + 7 = (1).
Here JA'' + B^= ^TTs = sJ4:=2.
Dividing (1) by 2, we have
i.e. ^(-i)+y(--^)-i=o,
i.e. X cos 240° + y sin 240° - 1 = 0.
58. To trace the straight line given hy an equation of
the first degree.
Let the equation be
Ax + By + G = (1).
(a) This can be written in the form
A B
Comparing this with the result of Art. 50, we see that it
(J
represents a straight Hne which cuts off intercepts — -^ and
— — from the axes. Its position is therefore known.
jO
If G be zero, the equation (1) reduces to the form
A
and thus (by Art. 47, Cor.) represents a straight hne
passing through the origin inchned at an angle tan~^ I ~ r )
to the axis of x. Its position is therefore known.
(^) The straight line may also be traced by firnding
the coordinates of any two points on it.
G
If we put y — in (1) we have x — —-r. The point
JL
i-'i-')
therefore lies on it.
38
COORDINATE GEOMETRY.
G
If we put oj = 0, we have 2/ = — ^ , so that the point
G^
(»-.)
lies on it.
line.
Hence, as before, we have the position of the straight
69. Ex. Trace the straight lines
(1) 3a;-4i/ + 7 = 0; (2) 7a; + 8y + 9 = 0j
(3) %y = x', (4) x = ^i (5) 2/= -2.
(1) Putting 2/ = 0, we have rc= -|,
and putting x = Q, we have y = ^.
Measuring 0A-^{= -^) along the axis of x we have one point on
the Hne.
Measuring OB^ (=t) along the axis of y we have another point.
Hence A-^B^ , produced both ways, is the required line,
(2) Putting in succession y and x equal to zero, we have the
intercepts on the axes equal to - f and - f.
If then 0-42= -f and 0^2= - |, we have A^B^, the required line.
(3) The point (0, 0) satisfies the equation so that the origin is on
the line.
Also the point (3, 1), i.e. C.^, lies on it. The required line is
therefore OC3.
(4) The line ic = 2 is, by Art. 46, parallel to the axis of y and passes
through the point A^ on the axis of x such that 0A^ = 2.
(5) The line y= - 2 is parallel to the axis of x and passes through
the point B^ on the axis of y, such that 0B^= - 2.
60. Straight Line at Infinity. We have seen
that the equation Ax + By + (7 = represents a straight line
STRAIGHT LINE JOINING TWO POINTS. 39
c c
which cuts oiF intercepts — - and — — from the axes of
Ji. Jj
coordinates.
If A vanish, but not B or C, the intercept on the axis
of X is infinitely great. The equation of the straight line
then reduces to the form y = constant, and hence, as in
Art. 46, represents a straight line parallel to Ox.
So if B vanish, but not A or C, the straight line meets
the axis of y at an infinite distance and is therefore parallel
to it.
If A and B both vanish, but not C, these two in-
tercepts are both infinite and therefore the straight line
Q .x + .y + C = is altogether at infinity.
61. The multiplication of an equation by a constant
does not alter it. Thus the equations
2a;-32/+5 = and 10a;- 152/+ 25 -
represent the same straight line.
Conversely, if two equations of the first degree repre-
sent the same straight line, one equation must be equal to
the other multiplied by a constant quantity, so that the
ratios of the corresponding coefficients must be the same.
For example, if the equations
a^x + \y + Ci = and A-^^x + B^y + Cj =
we must have
«! \ CjL
62. To jind the equation to the straight line which
passes through the two given points {x\ y') and (x", y").
By Art. 47, the equation to any straight line is
y-- mx -VG (1).
By properly determining the quantities m and c we can
make (1) represent any straight line we please.
If (1) pass through the point (a;', y')^ we have
2/' = mas' + c (2).
Substituting for c from (2), the equation (1) becomes
y-y' = m(x-x') (3).
40 COOKDINATE GEOMETRY.
This is the equation to the line going through (x\ y') making
an angle tan~^ m with OX. If in addition (3) passes through
the point {x", y"), then
y —y=m{x — x),
ti r
y -y
* * X' - X
Substituting this value in (3), we get as the required
equation
V" — v'
*^ X" — x^ '
63. Ex. Find the equation to the straight line which passes
through the points (-1, 3) and (4, -2).
Let the required equation be
y=mx + c (1).
Since (1) goes through the first point, we have
3=-m + c, so that c = m + S.
Hence (1) becomes
y = mx + m + S (2).
If in addition the line goes through the second point, we have
-2 = 47?i + m + 3, so that m= -1.
Hence (2) becomes
y=-x + 2, i.e. x + y = 2.
Or, again, using the result of the last article the equation is
y-B = ^-^^^{x + l)=-x-l,
i.e. y + x-=2.
64. To fix definitely the position of a straight line we
must have always two quantities given. Thus one point
on the straight line and the direction of the straight line
will determine it; or again two points lying on the straight
line will determine it.
Analytically, the general equation to a straight line
will contain two arbitrary constants, which will have to be
determined so that the general equation may represent any
particular straight line.
Thus, in Art. 47, the quantities m and c which remain
the same, so long as we are considering the same straigld
line, are the two constants for the straight line.
EXAMPLES. 41
Similarly, in Art. 50, the quantities a and h are the
constants for the straight line.
65. In any equation to a locus the quantities x and y,
which are the coordinates of any point on the locus, are
called Current Coordinates ; the curve may be conceived as
traced out by a point which " runs " along the locus.
EXAMPLES. V.
Find the equation to the straight line
1. cutting off an intercept unity from the positive direction of the
axis of y and inclined at 45° to the axis of x.
2. cutting off an intercept - 5 from the axis of y and being equally
inclined to the axes.
3. cutting off an intercept 2 from the negative direction of the
axis of y and inclined at 30° to OX.
4. cutting off an intercept - 3 from the axis of y and inclined at
an angle tan~i f to the axis of x.
Find the equation to the straight line
5. cutting off intercepts 3 and 2 from the axes.
6. cutting off intercepts - 5 and 6 from the axes.
7. Find the equation to the straight line which passes through the
point (5, 6) and has intercepts on the axes
(1) equal in magnitude and both positive,
(2) equal in magnitude but opposite in sign.
8. Find the equations to the straight lines which pass through
the point (1, - 2) and cut off equal distances from the two axes.
9. Find the equation to the straight line which passes through
the given point {x\ y') and is such that the given point bisects the
part intercepted between the axes.
10. Find the equation to the straight line which passes through
the point ( - 4, 3) and is such that the portion of it between the axes
is divided by the point in the ratio 5 : 3.
Trace the straight lines whose equations are
11. a; + 2?/+3 = 0. 12. 5a--7//-9 = 0.
13. 3a; + 7r/ = 0. 14. 2a;-3?/ + 4 = 0.
Find the equations to the straight lines passing through the
following pairs of points.
15. (0, 0) and (2, -2). 16. (3, 4) and (5, 6).
17. (-1, 3) and (6, -7). 18. (0, -a) and (&, 0).
42
COORDINATE GEOMETRY.
[Exs. v.]
19. (a, &) and {a + h, a-h).
20. {at^, 2at-^) and (at^^ 2at;).
21. (a«„^)and(a«„^j.
22. (« cos 01 , a sin <pi) and (a cos (p^, a sin ^a)-
23. (acos0jLJ & sin 0j) and (acos02> ^sin^g)*
24. (* sec 01, 6 tan 0i) and (a sec 02, 6 tan 02).
Find the equations to the sides of the triangles the coordinates of
whose angular points are respectively
25. (1,4), (2,-3), and (-1,-2).
26. (0,1), (2,0), and (-1, -2).
27. Find the equations to the diagonals of the rectangle the
equations of whose sides are x = a, x = a\ y = b, and y = b\
28. Find the equation to the straight line which bisects the
distance between the points {a, b) and {a', b') and also bisects the
distance between the points ( - a, b) and (a', - b').
29. Find the equations to the straight lines which go through the
origin and trisect the portion of the straight line 3a; + 2/ = 12 which
is intercepted between the axes of coordinates.
Angles between straight lines.
66. To find the angle between two given straight lines.
Let the two straight lines be AL^ and AL^j meeting the
axis of X in L^ and L^,
I. Let their equations be
y — m^x^-G-^ and y ~ in.j,x ^r c.^
By Art. 47 we therefore have
tan^ZjA'^mi, and td^Vi. AL.^X^Wj.^,.
Now L L-^AL^^ — L AL^X — L AL.2.X.
tan L^AL^ — tan \AL^X — AL^X\
ta,n AL^X— tan AL^X rn^ — n^
(1).
1 + tan AL^X. tan AL^X
1 +mi«i2
ANGLES BETWEEN STRAIGHT LINES. 43
Hence the required angle — lL^AL
= tan-i "'^•""'^ (2).
l + mim2
[In any numerical example, if the quantity (2) be a positive quan-
tity it is the tangent of the acute angle between the lines ; if negative,
it is the tangent of the obtuse angle.]
II. Let the equations of the straight lines be
^i£c + ^i2/ + Ci = 0,
and A^^x^- B^^y + G^^O.
By dividing the equations by B^ and B^, they may be
written
and
A,
c,
2/--
A.
'A'
L
y = -
sr
A'
Comparing
these with the equations of
(I-x
we
see
that
.„ ^1
A
^2
Hence the required angle
- tan-i-,— i == tan ^
B
r(-J)
1 + mi7n.2 , / -41"^^ ^ -^2^
(-!)(-»
= *"" 3STA^ <^>-
III. If the equations be given in the form
X cos a + y sin a — ^^ = and x cos ^ + 2/ sin j^ —p-^ — O,
the perpendiculars from the origin make angles a and p
with the axis of x.
Now that angle between two straight lines, in which
the origin lies, is the supplement of the angle between the
perpendiculars, and the angle between these perpendiculars
is ^ — a.
[For, if OEi and OR2 be the perpendiculars from the origin upon
the two hnes, then the points O, R^, R^, and A lie on a circle, and
hence the angles R^OR^ and R.^AR^ are either equal or supplementary.]
44 COORDINATE GEOMETRY.
67. To find the condition that two straight lines may
he parallel.
Two straight lines are parallel when the angle between
them is zero and therefore the tangent of this angle is zero.
The equation (2) of the last article then gives
Two straight lines whose equations are given in the
"m" form are therefore parallel when their "7?i's" are the
same, or, in other words, if their equations differ only in
the constant term.
The straight line Ax + By + G' = is any straight line which is
parallel to the straight line Ax + By + C = 0. For the "m's" of the
two equations are the same.
Again the equation A {x-x')+B {y-y') = clearly represents the
straight line which passes through the point {x', y') and is parallel to
Ax + By + C=0.
The result (3) of the last article gives, as the condition
for parallel lines,
68. Ex. Find the equation to the straight line, which passes
through the point (4, - 5), and which is parallel to the straight line
3:c + 4r/ + 5--=0 (1).
Any straight line which is parallel to (1) has its equation of the
form
3a; + 4^/ + (7=0 (2).
[For the "w" of both (1) and (2) is the same.]
This straight line will pass through the point (4, - 5) if
3x4 + 4x(-5) + C = 0,
i.e. if (7=20-12 = 8.
The equation (2) then becomes
3a;+42/ + 8 = 0.
69. To find the condition that two st^'aight lines j whose
equations are given, may he 'perpendicular.
Let the straight lines be
y — m^x -i-Ci,
and y — m.^x -\- G.2_.
CONDITIONS OF PERPENDICULARITY. 45
If be the angle between them we have, by Art. 66,
tan^^ r^""^^ (1).
1 +mim2
If the lines be perpendicular, then ^ = 90°, and therefore
tan = 00 .
The right-hand member of equation (1) must therefore
be infinite, and this can only happen when its denominator
is zero.
The condition of perpendicularity is therefore that
1 + m^TTi^ — O, i.e. Tn-^Tn2 = — I.'
The straight line y — tu^x + c.^ is therefore perpendicular
to y = »...H-.c., if «, = -!.
y/c'-t
It follows that the straight lines
A^x +B^y + C^ = and A^x + B^y + 0^ = 0,
for which m^ = — ^ and m^^ — ^ , are at right angles if
a) V A
AA / A,, _
i.e. a A^A^+B^B^ = 0.
70. From the preceding article it follows that the two
straight lines
A^x + B,y + Ci = Q (1),
and B,x-A,y+C^ = (2),
are at right angles ; for the product of their m's
Also (2) is derived from (1) by interchanging the coefficients
of a; and y, changing the sign of one of them, and changing
the constant into any other constant.
Ex. The straight line through (x', y') perpendicular to (1) is (2)
where B^x' - A-^y' + 62= 0, so that Cg = A^y'- B^x'.
This straight line is therefore
B,{x-x')-A^{y-y') = 0.
46 COORDINATE GEOMETRY.
71. Ex. 1. Find the equation to the straight line which passes
through the point (4, —5) and is perpendicular to the straight line
Sx + 4ij + 5 = (1).
First Method. Any straight line perpendicular to (1) is by the
last article
4:X-Sij + C=0 (2).
[We should expect an arbitrary constant in (2) because there are
an infinite number of straight lines perpendicular to (1).]
The straight line (2) passes through the point (4, - 5) if
4x4-3x(-5) + C = 0,
i.e. a (7= -16-15= -31.
The required equation is therefore
4:X-Sy = 31.
Second Method. Any straight line passing through the given
point is
y -{-5)=m{x~4:).
This straight line is perpendicular to (1) if the product of their
m's is - 1,
i.e. if m X ( - 1) = - 1,
i.e. if m=|.
The required equation is therefore
y + 5=i{x-4),
i.e. 4:X-'6y = Sl.
Third Method. Any straight line is y=mx + c. It passes through
the point (4, - 5), if
-5 = 4m + c (3).
It is perpendicular to (1) if
mx{-i)=-l (4).
Hence m = f and then (3) gives c = — V.
The required equation is therefore y = '^x-^-^,
i.e. 4x-By = Sl.
[In the first method, we start with any straight line which is
perpendicular to the given straight line and pick out that particular
straight line which goes through the given point.
In the second method, we start with any straight line passing
through the given point and pick out that particular one which is
perpendicular to the given straight hne.
In the third method, we start with any straight line whatever and
determine its constants, so that it may satisfy the two given
conditions.
The student should illustrate by figures. ]
Ex. 2. Find the equation to the straight line which passes through
the point (x', y') and is perpendicular to the given straight line
yy' = 2a {x + x').
THE STRAIGHT LINE. 47
The given straight line is
yy' - 2ax - 2ax' = 0.
Any straight line perpendicular to it is (Art. 70)
2ay + xy'+G=0 (1).
This will pass through the point (x', y') and therefore will be the
straight line required if the coordinates x' and y' satisfy it,
i.eAt 2ai/ + xY+C = 0,
i.e. if G=-2ay' -x'y'.
Substituting in (1) for G the required equation is therefore
2a{y-y') + y'{x-x') = 0.
72. To find the equations to the straight lines which
pass through a given point (x', t/') and make a given angle a
with the given straight line y — nix + c.
Let P be the given point and let the given straight line
be LMJSf, making an angle
with the axis of x such that
tan = m.
In general (i.e. except when
a is a right angle or zero) there
are two straight lines PMR and
FNS making an angle a with
the given line.
Let these lines meet the axis of x in R and S and let
them make angles ^ and <^' with the positive direction of
the axis of x.
The equations to the two required straight lines are
therefore (by Art. 62)
y -y' = tan ^ x (x — x) (1),
and y ~y' ~ t^^ <fi' X (x — x') (2).
Now cf, = L LMR + L RLM = a + 6*,
and <^'^L LNS+ L SLN= (180° ^a) + e.
Hence
■ , ^. tan a + tan ^ tana + m
tan <^ = tan (a + ^) - .j — ^ = ,
i — tan a tan ff i —m tan a
and tan </>' = tan ( 1 80° + ^ - a)
. „ . tan d — tan a m— tan a
= tan (^ — a) = ^— -p, — = ., .
1 + tan tan a 1 +7n tan a
48 COORDINATE GEOMETRY.
On substituting these values in (1) and (2), we have as
the required equations
, m + tan a , ,,
^ ^ 1-mtana^ ^
, m — tan a , ,.
and y-y = 1 1 \^ - ^ )•
^ ^ 1 + m tan a
EXAMPLES. VI.
Find the angles between the pairs of straight lines
1. x-ijsj^ — ^ and ^/3a;+2/ = 7.
2. ic-4?/ = 3 and ^x-y = ll. 3. 2/ = 3a3 + 7 and 3?/-a; = 8.
4. 2/ = (2-V3)a: + 5 and 2/= (2 + ^/3) a;- 7.
5. {m'^-mn)y = (inn^-n^)x + n^ and (?n7H- m^) y = (??i?i - w'^) a; + m^
6. Find the tangent of the angle between the lines whose inter-
cepts on the axes are respectively a, - 6 and 6, - a.
7. Prove that the points (2, - 1), (0, 2), (2, 3), and (4, 0) are the
coordinates of the angular points of a parallelogram and find the
angle between its diagonals.
Find the equation to the straight line
8. passing through the point (2, 3) and perpendicular to the
straight line 4a;-3i/ = 10.
9. passing through the point ( - 6, 10) and perpendicular to the
straight line 7aj + 8?/ = 5.
10. passing through the point (2, -3) and perpendicular to the
straight line joining the points (5, 7) and ( - 6, 3).
11. passing through the point (-4, -3) and perpendicular to the
straight line joining (1, 3) and (2, 7).
12. Find the equation to the straight line drawn at right angles to
the straight line — v = 1 through the point where it meets the axis
a b
of X.
13. Find the equation to the straight line which bisects, and is
perpendicular to, the straight line joining the points (a, b) and
(a', &')•
14. Prove that the equation to the straight line which passes
through the point {a cos^ 6, a sin^ 6) and is perpendicular to the
straight line xsecd + y cosec d = ais x cos d-y sin d = a cos 26.
15. Find the equations to the straight lines passing through {x', y')
and respectively perpendicular to the straight lines
xxf-\-yy'=a\
[Exs. VI.]
EXAMPLES.
49
XX yy
= 1,
a- 62
and x'y + xy' = a-.
16. Find the equations to the straight lines which divide, internally
and externally, the line joining (-3,7) to (5, - 4) in the ratio of 4 : 7
and which are perpendicular to this line.
17. Through the point (3, 4) are drawn two straight lines each
inclined at 45° to the straight line x-y = 2. Find their equations
and find also the area included by the three lines.
18. Shew that the equations to the straight lines passing through
the point (3, - 2) and incHned at 60° to the line
iJ3x + y = l are y + 2 = and y -J3x + 2 + 3^S = 0.
19. Find the equations to the straight lines which pass through
the origin and are inclined at 75° to the straight line
x + y + ^S{y-x) = a.
20. Find the equations to the straight lines which pass through
the point {h, k) and are inclined at an angle tan~'^m to the straight
line y = mx + c.
21. Find the angle between the two straight lines 3a; = 4?/ + 7 and
5y = 12x + 6 and also the equations to the two straight lines which
pass through the point (4, 5) and make equal angles with the two
given lines.
73. To sheiv that the point (x', y') is on one side or the
other of the straight line Ax + By +(7 = according as the
quantity Ax + By' + C is positive or negative.
Let LM be the given straight line and P any point
ix\ y).
Through P draw P^, parallel to
the axis of 3/, to meet the given
straight line in Q^ and let the co-
ordinates of Q be (.'«', y").
Since Q lies on the given line, we
have
^£c' + %" + C=:0,
Ax + C
y
i>x:x
so that
B
.(1).
It is clear from the figure that PQ is drawn parallel to
the positive or negative direction of the axis of y according
as P is on one side, or the other, of the straight line LM^
i.e. according as y" is > or < y\
i.e. according as y" — y is positive or negative.
L.
4
50 COORDINATE GEOMETKY.
Now, by (1),
The point {x ^ y') is therefore on one side or the other of
LM according as the quantity Ax' + By' + C is negative or
positive.
Cor. The point (cc', y') and the origin are on the same
side of the given line if Ax + By + G and AxO-\- B xO + C
have the same signs, i.e. if Ax' + By' + C has the same sign
as C.
If these two quantities have opposite signs, then the
origin and the point (x, y') are on opposite sides of the
given line.
!74. The condition that two points may lie on the
same or opposite sides of a given line may also be obtained
by considering the ratio in which the line joining the two
points is cut by the given line.
For let the equation to the given line be
Ax + By-¥C=0 (1),
and let the coordinates of the two given points be [x-^, y^
and (a?2, 2/2)-
The coordinates of the point which divides in the ratio
7?ii : vu the line joining these points are, by Art. 22,
Tn^ + TYi.j 7)1-^ + m^ ^ ''
If this point lie on the given line we have
^^ m^x^ + m^x^ ^ ^ m^y^ + m^y^ ^ ^^^
mj + mg m^ + m^ '
^1 . ^1 Ax.-\-By. + G /«,
so that _i = _^ w^—Ti (3 •
m^ Ax^ + ^2/2 + ^
If the point (2) be between the two given points {x-^, y^)
and (x^, 2/2), i.e. if these two points be on opposite sides of
the given line, the ratio in-^ : 711^ is positive.
In this case, by (3) the two quantities Ax^ + By^ + C
and Ax^ + By^ + C have opposite signs.
The two points (a?i, y^ and {x^^ y^) therefore lie on the op-
LENGTHS OF PEKPENDICULARS. 51
posite (or the same) sides of the straight line Ax + By + (7 =
according as the quantities Ax^ + By^ + G and Ax.^ + By.^ + G
have opposite (or the same) signs.
Lengths of perpendiculars.
75. To find the length of the perpendicular let fcdl from,
a given point upon a given straight line.
. Y
(i) Let the equation of the straight line be
£CCOSa+ 2/ sin a — p = (1),
so that, if p be the perpendicular on it, we have
ON^p and lXON=^o..
Let the given point F be {x ^ y').
Through P draw PR parallel to the given line to meet
OiV produced in R and draw PQ the required perpendicular.
If OR be 2^'i the equation to PR is, by Art. 53,
X cos a + 2/ ^^^ ^ ~P' — ^•
Since this passes through the point {x, y\ we have
£c' cos a + 2/' sin a.—p' = 0,
so that p = x' cos a + 2/' sin a.
But the required perpendicular
^PQ = NR = OR-ON = p'-p
= X' cos a + y' sin a — p (2).
The length of the required perpendicular is therefore
obtained by substituting x and y for x and y in the given
equation.
(ii) Let the equation to the straight line be
Ax + By + G=^0 (3),
the equation being written so that (7 is a negative quantity.
4—2
52 COORDINATE GEOMETRY.
As in Art. 56 this equation is reduced to the form (1)
by dividing it by V-^^ + B^. It then becomes
^A^-^B" slA^^B" ^|A^\E'
Hence
A , B ^ G
cos a = ,,„ , sm a = ,-- , and — «
^'A^ + B^' \/J'' + B ^A'^ + B''
The perpendicular from the point {x\ y') therefore
= x' cos a + 2/" sin a—p
_ Ax' + By^ + C
VAZ + B2
The length of the perpendicular from (x, y') on (3) is
therefore obtained by substituting x and y for x and 2/ in
the left-hand member of (3), and dividing the result so
obtained by the square root of the sum of the squares of
the coefficients of x and y.
Cor. 1. The perpendicular from the origin
^G-rs]~AFVB\
Cor. 2. The length of the perpendicular is, by Art. 73,
positive or negative according as {x ^ y) is on one side or
the other of the given line.
76. The length of the perpendicular may also be
obtained as follows :
As in the figure of the last article let the straight line
meet the axes in L and M^ so that
OL^ -% and 0M=-%,.
A B
Let PQ be the perpendicular from F (re', y'^ on the
given line and PS and PT the perpendiculars on the axes
of coordinates.
"We then have
/\PML + /\MOL = /\OLP + AOPM,
i.e., since the area of a triangle is one half the product of
its base and perpendicular height,
PQ,LM+OL. OM^ OL.PS+ OM . PT.
EXAMPLES. 53
since (7 is a negative quantity.
Hence
Ax' + By' +
so that P<?r=
Ja^ + b^
EXAMPLES. VII.
Find the length of the perpendicular drawn from
1. the point (4, 5) upon the straight line 3a; + 4?/ = 10.
2. the origin upon the straight line -^ - j=l.
3. the point ( - 3, - 4) upon the straight line
■12{x + 6) = 5{y-2).
4. the point (&, a) upon the straight line — f =!•
5. Find the length of the perpendicular from the origin upon the
straight line joining the two points whose coordinates are
{a cos a, a sin a) and (a cos j8, a sin j8).
6. Shew that the product of the perpendiculars drawn from the
two points ( ± \/a2 - h^, 0) upon the straight line
- cos ^ + ^ sin ^= 1 is 62.
a
7. If p and p' he the perpendiculars from the origin upon the
straight lines whose equations are x sec ^ + ^ cosec 6 = a and
a; cos ^ - 2/ sin ^ = a cos 2^,
prove that 4Lp^+p'^ = a'^.
8. Find the distance between the two parallel straight lines
y = mx + c and y = mx + d.
9. What are the points on the axis of x whose perpendicular
X 1/
distance from the straight line - + ~ =lis a^
ah
10. Shew that the perpendiculars let fall from any point of the
straight line 2a; + 11?/ = 5 upon the two straight lines 24a; + 7t/ = 20
and 4a; -3?/ = 2 are equal to each other.
54 COORDINATE GEOMETRY. [EXS. VII.l
11. Find the perpendicular distance from the origin of the
perpendicular from the point (1, 2) upon the straight line
77. To find the coordinates of the foint of intersection
of two given straight lines.
Let the equations of the two straight lines be
a-^x + h{y + Ci = (1),
and fta^^ + 522/ + ^2 = {2)5
and let the straight lines be AL^ and AL^ as in the figure
of Art. 66.
Since (1) is the equation of AL^^ the coordinates of any
point on it must satisfy the equation (1). So the coordi-
nates of any point on AL^ satisfy equation (2).
Now the only point which is common to these two
straight lines is their point of intersection A.
The coordinates of this point must therefore satisfy
both (1) and (2).
If therefore A be the point {x^, t/i)? ^^ have
«ia^i + ^i2/i + Ci = (3),
and a2^i + 522/i + <^2 = W-
Solving (3) and (4) we have (as in Art. 3)
^x ^ 2/1 ^ \__
so that the coordinates of the required common point are
h-fi^ — h.^c^ - c^a^ — c^a^
a^^-ajb^ a-}><^—a^^
78. The coordinates of the point of intersection found
in the last article are infinite if
a^^ — a^hi = 0.
But from Art. 67 we know that the two straight lines
are parallel if this condition holds.
Hence parallel lines must be looked upon as lines whose
point of intersection is at an infinite distance.
CONCUKEENCE OF STRAIGHT LINES. 55
79. To find the condition that three straight lines may
ineet in a point.
Let their equations be
a^x + hyy + c^ = (1),
a.jX + h^y + C2 = (2),
and a.p: + h-^y + C3 = (3).
By Art. 77 the coordinates of the point of intersection
of (1) and (2) are
—r -J- and —jf J-- (4).
If the three straight lines meet in a point, the point of
intersection of (1) and (2) must lie on (3). Hence the
values (4) must satisfy (3), so that
0-iCn — Oc\C-\ -t C-\Cto C.-^Cti -
«3 X -^r — T- + ^3 X -V — ~-Y + C3 = 0,
i. e. a^ (b^c^ — b^c-^) + 63 (ci^g — 02%) + c^ (a^b^ — a^bj) — 0,
i. e. ai (b^c^ — b^c^) + b^ {c^a^ — c^a^ + Cj (^2^3 — ^3^2) = . . , (5).
Aliter. If, the three straight lines meet in a point let
it be (ccj, 2/1), so that the values x\ and y-^ satisfy the
equations (1), (2), and (3), and hence
a-^x-^ + 6i2/i + Ci = 0,
ftoa^i + b^yi + C2 = 0,
and a^x-^ + b^y^ + Cg = 0.
The condition that these three equations should hold
between the two quantities x^ and y^ is, as in Art. 12,
^3 J ^3 ) <^3
which is the same as equation (5).
80. Another criterion as to whether the three straight
lines of the previous article meet in a point is the following.
If any three quantities ^p, q, and r can be found so
that
p (ajX + b^y + C;^) + q (a^x + b.2y + c^) + r (a^x + b^y + Cg) =
identically, then the three straight lines meet in a point.
56 COORDINATE GEOMETRY.
For in this case we have
a^x + h^y + Cg =: — - (a-^x + h^y + Cj) — - {a^x + h^y + Co) . • •(!).
Now the coordinates of the point of intersection of the
first two of the lines make the right-hand side of (1) vanish.
Hence the same coordinates make the left-hand side vanish.
The point of intersection of the first two therefore satisfies
the equation to the third line and all three therefore meet
in a point.
81. Ex. 1. Shew that the three straight lines 2«-3i/ + 5 = 0,
dx + ^y -1 = 0, and 9a; -5y 4- 8 = meet in a point.
If we multiply these three equations by 6, 2, and - 2 we have
identically
6 (2x - 32/ + 5) + 2 (3x -I- 4t/ - 7) - 2 (9a; - 5?/ -F 8) = 0.
The coordinates of the point of intersection of the first two lines
make the first two brackets of this equation vanish and hence make
the third vanish. The common point of intersection of the first two
therefore satisfies the third equation. The three straight lines
therefore meet in a point.
Ex. 2. Prove that the three 'perpendiculars draicn from the
vertices of a triangle upon the opposite sides all meet in a point.
Let the triangle be ABC and let its angular points be the points
K,2/i). (^2 '2/2)5 and (x^^ys)-
The equation to BG is y-y^ = "^ — - {x - x^).
The equation to the perpendicular from A on this straight line is
x^ x^
i'^' 2/ (2/3 -2/2) + ^(^3 -^2) =2/1 (1/3 -2/2) +^1(^3 -^2) (!)•
So the perpendiculars from B and C on CA and AB are
2/(2/i-2/3)+^(aa-^3) = 2/2(2/i-2/3)+«'2{^i-^3) (2).
and 2/ (2/2 -2/]) + ^(^2-^1) =2/3 (2/2 -2/1) + ^3 (^2-^1) (3).
On adding these three equations their sum identically vanishes.
The straight lines represented by them therefore meet in a point.
This point is called the ortliocentre of the triangle.
82. To Jind the equation to any straight line which
2)asses through the intersection of the two straight lines
a-^x + \y + c-^ = (1)>
and a^x + b^y + Cg = (2).
INTERSECTIONS OF STRAIGHT LINES. 57
If (a?i, 2/1) be the common point of the equations (1)
and (2) we may, as in Art. 77, find the values of x^ and 2/1,
and then the equation to any straight line through it is
where m is any quantity whatever.
Aliter. If A be the common point of the two straight
lines, then both equations (1) and (2) are satisfied by the
coordinates of the point A.
Hence the equation
a-^x + h^y + Ci + X {a,^x + h^y + Co) = (3)
is satisfied by the coordinates of the common point A,
where \ is any arbitrary constant.
But (3), being of the first degree in x and y, always
represents a straight line.
It therefore represents a straight line passing through A .
Also the arbitrary constant X may be so chosen that (3)
may fulfil any other condition. It therefore represents
any straight line passing through A.
83. Ex. Find the equation to the straight line ivhich passes
through the intersection of the straight lines
2x-Sy + 4: = 0, Sx + ^y-5 = (1),
and is perpendicular to the straight line
Gx-7y + 8 = (2).
Solving the equations (1), the coordinates x^^, 7/1 of their common
point are given by
^1 ^ yi _ 1 _ 1
(-3)(-5)-4x4 4x3-2x(-5) 2x4-3x(-3) ^^'
so that x-^= - yY and 2/1 =Tf-
The equation of any straight line through this common point is
therefore
y-H=m{x + ^).
This straight line is, by Art. 69, perpendicular to (2) if
m X f = - 1, i.e. if mr= - |.
The required equation is therefore
i.e. 119a; + 1022/ = 125.
58
COORDINATE GEOMETRY.
Aliter. Any straight line through the intersection of the straight
lines (1) is
i.e. (2 + 3X)a; + ?/(4X-3) + 4-5X = (3).
This straight line is perpendicular to (2), if
6(2 + 3X)-7(4X-3) = 0, (Art. 69)
i.e. a \ = u.
*
The equation (3) is therefore
i.e. 119a; +102?/ -125 = 0.
Bisectors of angles between straight lines.
84. To find the equations of the bisectors of the angles
between the straight lines
ajpc + b-yy + Cj^ = (1),
and a^ + b^ + c^ = (2).
Let the two straight lines be AL^ and AL^, and let the
bisectors of the angles between them be AM^ and A^f^-
Let P be any point on either of these bisectors and
draw PiVi and FJV^ perpendicular to the given lines.
The triangles PAN-^ and PAN<^ are equal in all respects,
so that the perpendiculars PN^ and PN^ ^-re equal in
magnitude.
Let the equations to the straight lines be written
so that Ci and c^ are both negative, and to the quantities
Ja^ + b^ and Ja^ + b^ let the positive sign be prefixed.
EQUATIONS TO BISECTOKS OF ANGLES. 59
If P be the point (/i, k), the numerical values of PN^
and PN^ are (by Art. 75)
aJh + hJc + c. , aji + hjc + c^ , ,
— ,_ and — . (1).
If P lie on AM^^ i.e. on the bisector of the angle
between the two straight lines in which the origin lies, the
point P and the origin lie on the same side of each of the
two lines. Hence (by Art. 73, Cor.) the two quantities (1)
have the same sign as c^ and c^ respectively.
In this case, since c^ and c^ have the same sign, the
quantities (1) have the same sign, and hence
aji + hjc + Ci aji + hjc + c^
But this is the condition that the point (h, k) may lie on
the straight line
a^x + h-{y + Cj a^ + h^y + c^
which is therefore the equation to AM-^.
If, however, P lie on the other bisector AM^, the two
quantities (1) will have opposite signs, so that the equation
to AM^ will be
a-^x + \y + Ci a^x + })c^y + c^
The equations to the original lines being therefore
arranged so that the constant terms are both positive (or
both negative) the equation to the bisectors is
VS7+b7 - Vi7+bp '
the upper sign giving the bisector of the angle in which
the origin lies.
85. Ex. Find the equations to the bisectors of the angles
between the straight lines
3x-4:y + 7 = and 12x-5y-S = 0.
Writing the equations so that their constant terms are both
positive they are
Sx-Ay + 7 = and -12x + 5y + 8 = 0.
60
COORDINATE GEOMETRY.
t.e.
i.e.
The equation to the bisector of the angle in which the origin lies
is therefore
Bx-^y + 7 _ -12x + 5y + 8
i.e. 13{Bx-^y + l) = 5{-12x + 5y + 8),
i.e. 99a; -772/ + 51 = 0.
The equation to the other bisector is
Sx-iy + l _ - 12x + 5?/ + 8
V32 + 42 ~ ~ x/122 + 52 '
13 {3a; - 4?/ + 7) + 5 ( - 12a; + 5?/ + 8) = 0,
21a: + 272/- 131 = 0.
86. It will be found useful in a later chapter to have
the equation to a straight line, which passes through a
given point and makes a given angle 6 with a given line, in
a form different from that of Art. 62.
Let A be the given point {h, k) and L'AL a straight
line through it inclined at an
angle 6 to the axis of x.
Take any point F, whose
coordinates are (x, y), lying on
this line, and let the distance
AP be r.
Draw PM perpendicular
to the axis of x and AN perpendicular to PM.
Then x--h=- AN = AP cosO=r cos 6,
and y — k = NP - AP sin ^ = r sin 0.
x-h y-k
Hence
= r
.(1).
This being the relation holding between the coordinates
of any point P on the line is the equation required.
Cor. From (1) we have
x — h + r cos 6 and y = k + r sin 0.
The coordinates of any point on the given line are
therefore h-\-r cos 6 and k + r sin 0.
87. To find the length of the straight line drawn
through a given point in a given direction to meet a given
straight line.
EXAMPLES. 61
Let the given straight line be
Ax-¥By + C^Q (1).
Let the given point A be (7t, k) and the given direction
one making an angle 6 with the axis of x.
Let the line drawn through A meet the straight line
(1) in P and let AP be r.
By the corollary to the last article the coordinates
of P are
h + r cos 6 and k + r sin 6.
Since these coordinates satisfy (1) we have
^ (A + r cos ^) + ^ (^ + r sin ^) + (7 - 0.
Ah-[-Bk+C
r =
(2),
A cos 6 ■¥ B sin 6
giving the length AP which is required.
Cor. From the preceding may be deduced the length
of the perpendicular drawn from (A, k) upon (1).
For the "m" of the straight line drawn through A is
tan ^ and the "m" of (1) is - -.
This straight line is perpendicular to (1) if
tan y. { — ^\ = — \,
i. e. if tan ^ = — ■ ,
so that
and hence
A
cos 6 sin 6 1
■4 B JA'- + B'
AcosO + BsinO= 4=^£L^s/A^ + B'.
V-i' + ^-
Substituting this value in (2) we have the magnitude
of the required perpendicular.
EXAMPLES. VIII.
Find the coordinates of the points of intersection of the straight
lines whose equations are
1. 2x-Si/ + 5 = and 7x-¥^y=S.
62 COOEDINATE GEOMETRY. [EXS.
2. - + T=1 and -+^ = 1.
a ha
3 a ^ a
. y = vi-,x-\ and y=moX-\ .
4. a; cos 01 + ^ sin 01 = a and a; cos 02+2/ sin 02 = «•
5. Two straight lines cut the axis of x at distances a and - a and
the axis of y at distances & and 6' respectively ; find the coordinates
of their point of intersection.
6. Find the distance of the point of intersection of the two
straight lines
2x-Sy + 5 = and dx + 4:y =
from the straight line
5x-2y = 0.
7. Shew that the perpendicular from the origin upon the
straight line joining the points
(a cos a, a sin a) and {a cos /3, a sin ^)
bisects the distance between them.
8. Find the equations of the two straight lines drawn through
the point (0, a) on which the perpendiculars let fall from the point
(2a, 2a) are each of length a.
Prove also that the equation of the straight line joining the feet
of these perpendiculars is y -r2x = oa.
9. Find the point of intersection and the inclination of the two
lines
Ax + By = A+B and A{x-y)+B{x + y)=2B.
10. Find the coordinates of the point in which the line
2y-Sx + l =
meets the line joining the two points (6, - 2) and ( - 8, 7). Find also
the angle between them.
11. Find the coordinates of the feet of the perpendiculars let fall
from the point (5, 0) upon the sides of the triangle formed by joining
the three points (4, 3), (-4, 3), and (0, -5); prove also that the
points so determined lie on a straight line.
12. Find the coordinates of the point of intersection of the
straight lines
2x-3y=^l and 5y-x = S,
and determine also the angle at which they cut one another.
13. Find the angle between the two lines
Sx + y + 12 = and x + 2y-l = 0.
Find also the coordinates of their point of intersection and the
equations of lines drawn perpendicular to them from the point
(3, -2).
VIII.] EXAMPLES. 63
14. Prove that the points whose coordinates are respectively
(5, 1), (1, -1), and (11, 4) lie on a straight line, and find its intercepts
on the axes.
Prove that the following sets of three lines meet in a point.
15. 2x-Sy = 7, Sx-4:y = 13, and 8x-lly = S3.
16. dx + 4.y + G = 0, 6x + 5y + 9 = 0, and Sx + Sy + 5 = 0.
17. - + 7 = 1, j+^ = l, and y = x.
abba
18. Prove that the three straight lines whose equations are
15a;- 18?/ + 1 = 0, 12x + lOi/ - 3 = 0, and 6x + QQy-ll =
all meet in a point.
Shew also that the third line bisects the angle between the other
two.
19. Find the conditions that the straight lines
y = m-^x + ai, y = m^-\-a^, and y = m2X-\-a^
may meet in a point.
Find the coordinates of the orthocentre of the triangles whose
angular points are
20. (0,0), (2, -1), and (-1,3).
21. (1,0), (2,-4), and (-5,-2).
22. In any triangle ABG^ prove that
(1) the bisectors of the angles A, B, and C meet in a point,
(2) the medians, i.e. the lines joining each vertex to the middle
point of the opposite side, meet in a point,
and (3) the straight lines through the middle points of the sides
perpendicular to the sides meet in a point.
Find the equation to the straight line passing through
23. tlie point (3, 2) and the point of intersection of the lines
2x + Sy = l and Sx-Ay = Q.
24. the point (2, - 9) and the intersection of the lines
2x + 5y-8 = and 3x-4y=^S5.
25. the origin and the point of intersection of
x~y-4i=0 and lx + y + 20=0,
proving that it bisects the angle between them.
26. the origin and the point of intersection of the lines
X y ^ ^ X y ^
- + f = 1 and Y + ^ = 1.
a b b a
27. the point (a, b) and the intersection of the same two lines.
28. the intersection of the lines
x-2y-a=0 and x + 3y-2a =
64 COORDINATE GEOMETRY. [Exs.
and parallel to the straight line
29. the intersection of the lines
x + 2y + S = and 3x + iy + 7 =
and perpendicular to the straight line
y-x = 8.
30. the intersection of the lines
dx-iy + l = and 5x + y -1=0
and cutting off equal intercepts from the axes.
31. the intersection of the lines
2x~By = 10 and x + 2y:=Q
and the intersection of the lines
16a;-102/ = 33 and 12x + Uy + 29 = 0.
32. If through the angular points of a triangle straight lines be
drawn parallel to the sides, and if the intersections of these Hnes be
joined to the opposite angular points of the triangle, shew that the
joining lines so obtained will meet in a point.
33. Find the equations to the straight lines passing through the
point of intersection of the straight lines
Ax + By + C = and A'x + B'y + C'^0 and
(1) passing through the origin,
(2) parallel to the axis of y,
(3) cutting off a given distance a from the axis of y,
and (4) passing through a given point {x', y').
34. Prove that the diagonals of the parallelogram formed by the
four straight lines
^?>x + y = 0, ^?>y + x=^0, jBx + y = l, and JBy + x = l
are at right angles to one another.
35. Prove the same property for the parallelogram whose sides
are
- + 7=1, r + - = l, - + | = 2, and t + - = 2.
a b a a o a
36. One side of a square is inclined to the axis of x at an angle a
and one of its extremities is at the origin ; prove that the equations
to its diagonals are
y (cos a - sin a) = x (sin a + cos a)
and ?/ (sin a + cos a) + a: (cos a -sin a) = a.
Find the equations to the straight lines bisecting the angles
between the following pairs of straight lines, placing first the bisector
of the angle in which the origin lies.
37. x+ysJB = %-\-2JB and a;-?/ ^3 = 6-2^3.
VIII.] EXAMPLES. 65
38. 12x + 5y-4c = and Bx+4.tj + 7 = 0.
39. ix + Sij -7 = and 24^ + 7ij- 31 = 0.
40. 2x + y=4: and y + Sx = 5.
41. y-b=^ i>(^-«) and y-h = z ^(rc-a).
^ l-m2^ ' ^ 1-m'^^ '
Find the equations to the bisectors of the internal angles of the
triangles the equations of whose sides are respectively
42. 3x + 4:y=e, 12x-5y=B, and 4:X-3y + 12 = 0.
43. Sx + 5y=15, x + y=4:, and 2x + y = Q.
44. Find the equations to the straight lines passing through the
foot of the perpendLcular from the point {h, Jc) upon the straight line
Ax + By + G = and bisecting the angles between the perpendicular
and the given straight line.
45. Find the direction in which a straight Kne must be drawn
through the point (1, 2), so that its point of intersection with the line
x + y = 4: may be at a distance ^^6 from this point.
CHAPTER V.
THE STRAIGHT LINE {continued).
POLAR EQUATIONS. OBLIQUE COORDINATES.
MISCELLANEOUS PROBLEMS. LOCI.
88. To find the general equation to a straight line in
polar coordinates.
Let p be the length of the perpendicular Y from the
origin upon the straight line, and
let this perpendicular make an
angle a with the initial line.
Let P be any point on the
line and let its coordinates be r
and 6.
The equation required will
then be the relation between r, 6, p, and a.
From the triangle YP we have
p = r cos YOP = rcos{a-6)=^r cos (6 - a).
The required equation is therefore
r cos (6 — a) =p.
[On transforming to Cartesian coordinates this equation becomes
the equation of Art. 53.]
89. To find the polar equation of the straight line
joining the poiiits whose coordinates are (r^, 6^) and {r^, 6^.
THE STRAIGHT LINE. OBLIQUE COORDINATES. 67
Let A and B be the two given points and P any point
on the line joining them
whose coordinates are r and
e.
Then, since
AA0B-=AA0P+A POB,
we have
^.e.
I.e.
J r^r^ sin A OB = J r^r sin AOP + ^ ri\ sin POB,
r^r^ sin {6 2 — 6^ = r^r sin {0 — 6-^ + rr^ sin (0^ — 6),
sin (^,-^i) _ sin ((9-^1) sin ((92-^)
OBLIQUE COORDINATES.
90. In the previous chapter we took the axes to be
rectangular. In the great majority of cases rectangular
axes are employed, but in some cases oblique axes may be
used with advantage.
In the following articles we shall consider the proposi-
tions in which the results for oblique axes are different
from those for rectangular axes. The propositions of Arts.
50 and 62 are true for oblique, as well as rectangular,
coordinates.
91. To find the equation to a straight line referred to
axes inclined at an angle w.
Let LPL' be a straight line which cuts the axis of
a distance c from the origin and is
inclined at an angle to the axis
of X.
Let P be any point on the
straight line. Draw PNM parallel
to the axis of y to meet OX in M^
and let it meet the . straight line
through C parallel to the axis of x
in the point N,
Let P be the point (.r, y\ so that
CN^OM^x, and NP = MP- 00 ^y-c.
5-
Fat
68 COORDINATE GEOMETRY.
Since L GPN= l FNN' - l PCN' = w - ^, we have
y-c NP _ B>inNC P _ sinO
~ir ~ 'CN~ ^^PN~ sin (o)- ^) •
TT Si^^ /1\
Hence y = x-. — ; 7:. + c (1).
^ sm(o>-^) ^ ^
3?his equation is of the form
y = mx + c,
where
sin^ sin^ tan^
7)1 =
sin (o) — 6) sin w cos — cos to sin sin <o — cos w tan ^ '
, , „ J /> ^ sin CD
and thererore tan v = .
1 + 7)1 cos a>
In oblique coordinates the equation
y = mx + c
therefore represents a straight line which is inclined at an
angle
- m sin o)
tan~i
l+mcosco
to the axis of x.
Cor. From. (1), by putting in succession equal to 90°
and 90° + o), we see that the equations to the straight lines,
passing through the origin and perpendicular to the axes of
X and y, are respectively y = and y = —x cos w.
92. The axes being oblique, to find the equation to the
straight line, such that the 'perpendicular on it from the origin
is of length p and makes angles a and ^ with the axes of x
and y.
Let LM be the given straight line and OK the perpen-
dicular on it from the origin.
Let P be any point on the
straight line ; draw the ordinate
PN and draw NP perpendicular
to OK and PS "oerpendicular to
NR.
Let P be the point {x, y), so
that OJV = X and NP = y.
THE STRAIGHT LINE. OBLIQUE COORDINATES. 69
The lines NP and Y are parallel.
Also OK and SP are parallel, each being perpendicular
to NB.
Thus lSPN^lKOM=^.
We therefore have
'p = OK- OR + SP = OiVcos a + NP cos p=^xcosa + y cos ^.
Hence x cos a + y cos /5 — p = 0,
being the relation which holds between the coordinates of
any point on the straight line, is the required equation.
93. To find the angle between the straight lines
y = mx + c and y = mx + c,
the axes being oblique.
If these straight lines be respectively inclined at angles
and 0' to the axis of x, we have, by the last article,
^ msinoD , ^ ,,, m'sinw
tan u = :. and tan u —
1 + ni cos CO 1 + 711 cos 0)
The angle required is 0~ 0'.
XT J. ir\ t\t\ tan ^- tan ^'
Now tan(e-e)=j^:^^^^^-^^,
m sm CD in sm co
1 +m cos CO 1 + 771 cos CO
m sin CO m' sin co
1 +
1+771 cos CO 1 + m' cos CO
_ m sin CO ( 1 + m' cos co) — m' sin co (1 + 7?i. cos co)
(1 + m cos co) (1 + W cos co) + ?92m' sin^ co
_ (m — w') sin CO
1 + (m + m ) cos CO + mm' '
The required angle is therefore
tan
_j (m — in!) sin co
1 + {m + TTb) cos 0) + 7f}im' '
Cor. 1. The two given lines are parallel if m = m'.
Cor. 2. The two given lines are perpendicular if
1 + (m + m') cos 0} + mm' = O.
70 COORDINATE GEOMETRY.
94. If the straight lines have their equations in the
form
Ax-\-By + G = and A'x + B'y ■\- C = 0,
then 7n = - ^ and m =--Wt'
Substituting these values in the result of the last article
the angle between the two lines is easily found to be
J A'B-AB' .
AA' + BB' - {AB' + A'B) cos w
The given lines are therefore parallel if
A'B-AB'=^0.
They are perpendicular if
AA' + BB' = {AB' + A'B) cos w.
95. Ex. The axes being inclined at an angle of 30°, obtain the
equations to the straight lines ivhich pass through the origin and are
inclined at 45° to the straight line x + y = l.
Let either of the required straight lines be y = mx.
The given straight line isy= -x + 1, so that m'= - 1.
We therefore have
1 + (m + m') cos <a + mm'
where m'= - 1 and w = 30°.
rw,- . ■■ ■ m+1 . -
This equation gives 2 + (,,,_ 1)^3- 2m = "^^^
Taking the upper sign we obtain m= — j^.
Taking the lower sign we have m = - ^3.
The required equations are therefore
y=-sjSx and y=z--^x,
i.e. y + iJ3x=^0 and JSy + x = 0.
96. To find the length of the lyerpendicular froin the
point (x'j y) upon the straight line Ax + By +0 — 0, the axes
being inclined at an angle <a, and the equation being written
so that C is a negative quantity.
THE STRAIGHT LINE. OBLIQUE COORDINATES. 71
Let the given straight line meet the axes in L and M^
G C
so that OL = --, and OM^ — -=, .
A B
Let P be the given point {x', y').
Draw the perpendiculars PQ, PR,
and PS on the given line and the
two axes.
Taking and P on opposite sides
of the given line, we then have
IiLPM + AMOL=aOLP + aOPM,
i.e. PQ . LM + OL . OM silicon OL . PE + OM . PS. ..{I).
Draw PU and PV parallel to the axes of y and x, so
that PU = y' and PV-^x.
Hence PE ^ PU sin PUR = y' sin w,
and PS =PV sin P VS - x sin w.
Also
LM= s/OL^ + OM^ - 20 L . Oif cos w
V
(72 C'
0^
z^-'^-^zg'"'"^
- (7 /l+l
2 cos w
since C is a negative quantity.
On substituting these values in (1), we have
pc><(-o)xy-i+-i
2 cos CO C^ .
c
c
so that
PQ =
--7.7/ sm o> — Pi . £c sm 00,
Ax^ + By' + C
VA2 + B2 - 2 AB cos CO
. sm a;.
Cor. If (0 — 90°, i.e. if the axes be rectangular, we
have the result of Art. 75.
72 COORDINATE GEOMETRY.
EXAMPLES. IX.
1. The axes being inclined at an angle of 60°, find the inclination
to the axis of x of the straight lines whose equations are
(1) 2/=2^ + 5,
and (2) 2y=.{^^-l)x + l.
2. The axes being inclined at an angle of 120°, find the tangent
of the angle between the two straight lines
Qx + ly=.l and 28a; - 73?/ = 101.
3. With oblique coordinates find the tangent of the angle
between the straight lines
y = mx + c and my+x = d.
4. liy=x tan -— - and y=x tan — j represent two straight lines
at right angles, prove that the angle between the axes is ~ .
5. Prove that the straight lines y + x=c and y=x + d are at
right angles, whatever be the angle between the axes.
6. Prove that the equation to the straight line which passes
through the point {h, Jc) and is perpendicular to the axis of x is
x + y cos o) = h+k cos w.
7. Find the equations to the sides and diagonals of a regular
hexagon, two of its sides, which meet in a corner, being the axes of
coordinates.
8. From each corner of a parallelogram a perpendicular is drawn
upon the diagonal which does not pass through that corner and these
are produced to form another parallelogram ; shew that its diagonals
are perpendicular to the sides of the first parallelogram and that they
both have the same centre.
9. If the straight lines y = miX + Cj^ and y=m,^x + C2 make equal
angles with the axis of x and be not pariallel to one another, prove
that iiij^ + ^2 + 2mjin2 cos w = 0.
10. The axes being inclined at an angle of 30°, find the equation
to the straight line which passes through the point ( - 2, 3) and is
perpendicular to the straight line y + Bx = 6.
11. Find the length of the perpendicular drawn from the point
(4, -3) upon the straight line 6a; + 3^ -10 = 0, the angle between the
axes being 60°.
12. Find the equation to, and the length of, the perpendicular
drawn from the point (1, 1) upon the straight line 3a; + 4?/ + 5 = 0, the
angle between the axes being 120°.
[EXS. IX.] THE STRAIGHT LINE. PROBLEMS. 73
13. The coordinates of a point P referred to axes meeting at an
angle w are [h, k) ; prove that the length of the straight line joining
the feet of the perpendiculars from P upon the axes is
sin w ^y/i^ +k- + 2hk cos w.
14. From a given point {h, k) perpendiculars are drawn to the
axes, whose inclination is oj, and their feet are joined. Prove that
the length of the perpendicular drawn from [h, Jc) upon this line is
hk sin^ 0}
JhF+W+2hkcos^'
and that its equation is hx - ky = h^- k^.
Straight lines passing through fixed points.
97. If the equation to a straic/ht line be of the form
ax + hy + c + \ (ax + b'y + g') = (1 ),
where \ is any arbitrary constant^ it always passes through
one fixed point whatever be the value of \.
For the equation (1) is satisfied by the coordinates of
the point which satisfies both of the equations
ax + by + G~Oj
and a'x 4- b'y + c' = 0.
This point is, by Art. 77,
'be' — b'c ca' — c'a^
^ab' — a'b ' ab' — a'b/
and these coordinates are independent of A.
Ex. Given the vertical angle of a triangle in magnitude and
position, and also the sum of the reciprocals of the sides ivhich contain
it; shew that the base always passes through a fixed point.
Take the fixed angular point as origin and the directions of the
sides containing it as axes ; let the lengths of these sides in any such
triangle be a and &, which are not therefore given.
We have - + 7=const. = r- (say) (1).
do K
The equation to the base is
^ y -r
— v- -=1
a
z...,by(l), ^'+^(1_1)^1,
1 y
i.e. -(x-y) + ^-l = 0.
74
COORDINATE GEOMETRY.
"Whatever be the value of a this straight line always passes through
the point given by
x-y = and |-1 = 0,
i.e. through ih.e fixed point {k, k).
k
98. Prove that the coordinates of the centre of the
circle inscribed in the triangle, whose vertices are the points
(^ij 2/i)> (^2 J 2/2)5 «^^G? (a^3, 2/3), are
axi + hx^ + cx^ ay^ + hy.2 + cy^
a+b+G a+b+c '
where a, 6, «7ic? c are ^Ae lengths of the sides of the triangle.
Find also the coordinates of the centres of the escribed
circles.
Let ABC be the triangle and let AD and CE be the
bisectors of the angles A and G
and let them meet in 0'.
Then 0' is the required point.
Since AD bisects the angle
BAC we have, by Euc. YI. 3,
^ _DG_ BD + DG _ _a_
BA ~ AG~ BA+'AC~bTc'
so that
ba
(xj.yj)
'(-,,y,)
DG
b + c
Also, since GO' bisects the angle AGD, we have
^' _ JLC _ _5_ _ 6 + c
0'D~ CD~ ba
a
b + <
The point D therefore divides BG in the ratio
BA : AG, i.e. c : b.
Also 0' divides AD in the ratio b + c : a.
Hence, by Art. 22, the coordinates of D are
cxs + bx^ fi ^3 + ^.'/2
c + b
c + b
THE STRAIGHT LINE. PROBLEMS.
Also, by the same article, the coordinates of 0' are
cxo + hxo ,, X cVo + hy.^
(b + c) X —^ — —^ + ax, (b + c)x -^ — ^^ + ay.
^ ' G+h T ' G+b ^
and
{b + c) + a (b + c) +
a
ax^ + bx^, + Gx.^ - ay^ 4- by^ + cy^
a + b + G a + b + c
Again, if 0^ be the centre of the escribed circle opposite
to the angle -4, the line COi bisects the exterior angle of
ACB.
Hence (Euc. VI. A) we have
AO, _AC__ b^G
Therefore Oi is the point which divides AD externally in
the ratio b + g : a.
Its coordinates (Art. 22) are therefore
,y . GXo + bx^ ,-, , cy., + by»
(b + c) ^ ^ ^ - ax^ (b + c) -^ — ^ - ay,
^ ^ G + b ^ , ^ ' c + b ^^
ana
{b + G) — a (^ + c)
a
— axi + bx^ + Gx^ - — ayi + by. 2 4- Gy^
—a+b+G —a+b+c
Similarly, it may be shewn that the coordinates of the
escribed circles opposite to B and C are respectively
^aXj^ — bx2 + GXs ay^ — by^ + cyA
c-
a—b+G ' a—b+c
and /axj^x2-_cxs ay^ + by.^ - Gy ^
\ a + b — c ' a + b — G
99. As a numerical example consider the case of the
triangle formed by the straight lines
3x+4cy-7 = 0, 12x + 6y-l7 =0 and 5x + 12y - 34: = 0.
These three straight lines being BC, CA, and AB
respectively we easily obtain, by solving, that the points
A, B, and C are
Q' t)' (t?' 11) ^"'^ (i'^>-
76 COORDINATE GEOMETRY,
Hence
yfi?-')"*(s-')'V
682 5p
17 ,j.-^, 85
-16^^"^^ ==16'
5=./ 1-^
and
?V /^l_i?V- A' 12-'_ 13
1) ^\T) ^ V r'^'r ~T"'
/ 72 52Y /19 _ 62Y _ /395T
V V7 "^ 16/ "^ VT~ W ~ V "~Tl2
165^
22
^^Vl69 '^'
112" 112'
Hence
85 2 170 85 19 1615
13 -52__676 13 67 871
429 , 429
c^z = YY9,> and C2/3 = Yj2-
The coordinates of the centre of the incircle are therefore
170 _ 6^6 m 1615 871 429
1X2 ~ iT2 "^ 112 , IT2" "^ 112 '^ II2
85 13 429 85 13 429'
16 ^T"^ 112 16 "^y^ 112
-1 ,265
^ and ^.
The length of the radius of the incircle is the perpen-
dicular from ( — T^ , jYo ) ^P^^ *^® straight line
3aj + 42/ - 7 = 0,
THE STRAIGHT LINE. PROBLEMS. 77
and therefore ^
i^)^0
-21 + 1060-784 255 51
5x112 5x112 112'
The coordinates of the centre of the escribed circle
which touches the side BG externally are
_170_676 429 1615 871 429
112 112^112 ~Tl2" "^ 112"^Tl2
_85 13 429 85 13 429 '
"le'^T'^m ~l6'^T'^ri2
-417 , -315
-42~ ^""^ -^r-
Similarly the coordinates of the centres of the other
escribed circles can be written down.
100. Ex. Find the radius, and the coordinates of the centre, of
the circle circumscribing the triangle formed by the points
(0, 1), (2, 3), and (3, 5).
Let (iCi , 2/j) be the required centre and R the radius.
Since the distance of the centre from each of the three points is the
same, we have
^i'+ (2/1 - 1)'= (^1 - 2)H (^/i - 3)2= (.ri - 3)2+ (y, - 5)2= J22...(l).
From the first two we have, on reduction,
^1 + 2/1 = 3.
From the first and third equations we obtain
Solving, we have x^= - ^ and yi=^.
Substituting these values in (1) we get
i2=tx/10.
101. Ex. Prove that the middle points of the diagonals of a com-
plete quadrilateral lie on the same straight line.
[Complete quadrilateral. Def. Let OAGB be any quadrilateral.
Let AG and OB be produced to meet in E, and BG and OA to meet in
F. Join AB, OC, and EF. The resulting figure is called a complete
quadrilateral ; the lines AB,OG, and EF are called its diagonals, and
the points E, F, and D (the intersection of AB and OC) are called its
vertices.]
78 COOEDINATE GEOMETRY.
Take the lines OAF and OBE as the axes of x and y.
1Y
BI ^
< v
/n>
\c -.
•.N
/ ^
* v^
n^X
[A
M-\
^^^
O A F X
Let 0A = 2a and 0B = 2b, so that A is the point (2a, 0) and B is
the point (0, 2b); also let C be the point {2h, 2k).
Then L, the middle point of OC, is the point {h, h), and ilf, the
middle point of AB, is (a, fo).
The equation to LM is therefore
£.e. {Ji-a)y-{k-h)x = bh-ak (1).
k-h
Again, the equation to BC is y -2b= - ~x.
Putting ?/ = 0, we have x = ^ — - , so that F is the point
f 2ak \
Similarly, E is the point ( 0, - _ j .
Hence N, the middle point of EF^ is f r; — r- , ■ j .
These coordinates clearly satisfy (1), i.e. N lies on the straight
line L3I.
EXAMPLES. X.
1. A straight line is such that the algebraic sum of the perpen-
diculars let fall upon it from any number of fixed points is zero;
shew that it always passes through a fixed point.
2. Two fixed straight lines OX and Y are cut by a variable line
in the points A and B respectively and P and Q are the feet of the
perpendiculars drawn from A and B upon the lines OBY and OAX.
Shew that, if ^B pass through a fixed point, then PQ will also pass
through a fixed point.
[EXS. X.] THE STRAIGHT LINE. PROBLEMS. 79
3. If the equal sides AB and AC of au isosceles triangle be pro-
duced to E and F so that BE .GF = AB% shew that the line EF will
always pass through a fixed point.
4. If a straight line move so that the sum of the perpendiculars
let fall on it from the two fixed points (3, 4) and (7, 2) is equal to
three times the perpendicular on it from a third fixed point (1, 3),
prove that there is another fixed point through which this line always
passes and find its coordinates.
Find the centre and radius of the circle which is inscribed in the
triangle formed by the straight lines whose equations are
5. 3a; + 42/ + 2 = 0, 3x-4.y + 12 = 0, and 4x-3y = 0.
6. 2x + iy + S = 0, 4x + Sy + 3 = 0, and a;+l = 0.
7. y = 0, 12x-5y=0, and 3a; + 4?/-7 = 0.
8. Prove that the coordinates of the centre of the circle inscribed
in the triangle whose angular points are (1, 2), (2, 3), and (3, 1) are
8 + v/lQ and ^^-^^^Q
g ana g .
Find also the coordinates of the centres of the escribed circles.
9. Find the coordinates of the centres, and the radii, of the four
circles which touch the sides of the triangle the coordinates of whose
angular points are the points (6, 0), (0, 6), and (7, 7).
10. Find the position of the centre of the circle circumscribing
the triangle whose vertices are the points (2, 3), (3, 4), and (6, 8).
Find the area of the triangle formed by the straight lines whose
equations are
11. y = x, y = 2x, and y = Sx + 4.
12. y + x=0, y=x + G, and y = 7x + 5.
13. 2y + x-5 = 0, y + 2x-7 = 0, and x-y + l = 0.
14. Sx-'iy + 4a = 0, 2x-By + 4a = 0, and 5x-y + a = 0, proving also
that the feet of the perpendiculars from the origin upon them are
coUinear.
15. y = ax-bc, y = bx-ca, and y = cx-db.
16. y — m.x-\ — , y = in^-\ , and y = moX + - — .
17. y=m^x + Ci, y — m^x + c^, and the axis oiy.
18. y=v\x + c-^, y =m^ + c^, and y=m^ + c^.
19. Prove that the area of the triangle formed by the three straight
lines a^x + h^y + c^ = 0, a^x + K^y + Cg = 0., and a^x + Z>3i/ + Cg = is
1 ^
2 -
^ 2
- -^ K^2 - «2&l) («2&3 - «3&2) («3&1 " ^1^ •
80 COORDINATE GEOMETRY. [ExS. X.]
20. Prove that the area of the triangle formed by the three straight
lines
X GOB a + y sin a- Pi = 0, xcos^ + y sin/S-^g^^?
and a; cos 7+?/ sin 7 -2)3 = 0,
^ sin (7 - /3) sin (a - 7) sin {^-a)
21. Prove that the area of the parallelogram contained by the
lines
4y-Sx-a=0, Sy -4:X + a = 0, 4:y-Sx-da=0,
and 3y-4:x + 2a=0 is faK
22. Prove that the area of the parallelogram whose sides are the
straight lines
aiX + biy + Ci = 0, ajX + bjy + dj^ = 0, a^x + b^y + c^^O,
and a^ + b2y + d2=0
IS
«1&2 - ^2^1
23. The vertices of a quadrilateral, taken in order, are the points
(0, 0), (4, 0), (6, 7), and (0, 3) ; find the coordinates of the point of
intersection of the two lines joining the middle points of opposite
sides.
24. The lines a; + 2/ + 1=0, x-y + 2=0, 4x + 2y + S=0, and
x + 2y-4: =
are the equations to the sides of a quadrilateral taken in order ; find
the equations to its three diagonals and the equation to the line on
which their middle points lie.
25. Shew that the orthocentre of the triangle formed by the three
straight lines
a a - a
y=mr.x-\ — , y=.m^x-\ — , and y=nux-\ —
is the point
\-a,a {— + — + — + U .
26. -4 and B are two fixed points whose coordinates are (8, 2) and
(5, 1) respectively ; ABP is an equilateral triangle on the side of AB
remote from the origin. Find the coordinates of P and the ortho-
centre of the triangle ABP.
102. £jX. The base of a triangle is fixed ; find the
locus of the vertex when one base angle is double of the
other.
THE STRAIGHT LINE. PROBLEMS.
81
Let AB be the fixed base of the triangle j take its
middle point as origin, the direc-
tion of 0£ as the axis of x and a
perpendicular line as the axis of y.
Let AO=OB^a.
If P be one position of the a O B
vertex, the condition of the problem then gives
lPBA^2^.PAB,
i.e. TT — cf> = 20,
i.e. —tan <^=:tan W (1).
Let P be the point (li, k). "We then have
= tan and -; = tan ^.
h + a h — a
Substituting these values in (1), we have
1{h + a)k
k
h + a
h
a
1
(h+ay-k^'
\h + a/
i.e. -{h + aY + ^=2{h^-a%
i.e. k''-3h^-2ah + a^^0.
But this is the condition that the point (A, k) should lie
on the curve
y^-3x'-2ax + a'' = 0.
This is therefore the equation to the required locus.
103. Ex. From a point P perpendiculars PM and
PN are draivn upon two fixed lines which are inclined at an
angle w and meet in a fixed point ; if P move on a fixed
straight line, find the locus of the middle point of MN.
Let the two fixed lines be taken as the axes. Let the
coordinates of P, any position of the
moving point, be {h, Tc).
Let the equation of the straight
line on which P lies be
Ax + By + (7 = 0,
so that we have
Ah + Bk + C = (1).
L.
82 COORDINATE GEOMETRY.
Draw PL and PL' parallel to the axes.
We then have
0M= OL + LM = OL + LP cos oi^h + Jc cos o>,
and 0J\^= OL' + L'N ^LP + L'P cosoi = k + h cos w.
M is therefore the point {h + k cos co, 0) and J^ is the point
(0, k + h cos co).
Hence, if {x, y) be the coordinates of the middle point
of JfiV, we have
2cc' =h-\-h cos CO (2),
and 2y —k + h cos co (3).
Equations (1), (2), and (3) express analytically all the
relations which hold between x, y\ A, and k.
Also li and k are the quantities which by their variation
cause Q to take up different positions. If therefore between
(1), (2), and (3) we eliminate h and k we shall obtain a
relation between x and y which is true for all values' of h
and k^ i. e. a relation which is true whatever be the position
that P takes on the given straight line.
From (2) and (3), by solving, we have
k ^ 2(a3^-y^cosco ) ^^^ ^ ^ 2 {y' - x' cos co) ^
sin^ 00 sin^ co
Substituting these values in (1), we obtain
2 A {x' - y' cos co) + 2^ {y — x cos co) + C sin^ co == 0.
But this is the condition that the point (cc', y) shall
always lie on the straight line
2A{x-y cos oi) + '2£ {y — X cos co) + C sin^ co - 0,
i. e. on the straight line
x{A — £ cos (j>) + y {£ — A cos co) + J C sin- co = 0,
which is therefore the equation to the locus of Q.
104. Ex. A straight line is drawn jjarallel to the
base of a given triangle and its extremities are joined trans-
versely to those of the base; find the locus of the 2>oint
of intersection of the joining lines.
THE STRAIGHT LINE. PROBLEMS.
83
Let the triangle be OAB and take as the origin and
the directions of OA and OB
as the axes of x and y.
Let OA = a and OB = 6,
so that a and h are given
quantities.
Let A'B' be the straight
line which is parallel to the
base ABj so that
OA ~~0B
and hence OA' = \a and OB' — Xb.
For different values of X we therefore have different
positions of A'B'.
The equation to AB' is
X (say),
and that to A'B is
- + — = 1
a Xb
(1).
(2)-
Since F is the intersection of AB' and A'B its coordi-
nates satisfy both (1) and (2). Whatever equation we
derive from them must therefore denote a locus going
through P. Also if we derive from (1) and (2) an equation
which does not contain X, it must represent a locus which
passes through P whatever be the value of A.; in other
words it must go through all the different positions of the
point P.
Subtracting (2) from (1), we have
l(}-l
^%(Ui
b\X
0,
X
%.e.
a b'
This then is the equation to the locus of P.
always lies on the straight line
b
y = — x.
Hence P
6—2
84 COORDINATE GEOMETRY.
which is the straight line OQ where OAQB is a parallelo-
gram.
Aliter. By solving the equations (1) and (2) we
easily see that they meet at the point
(:
\ri"' r?i*^-
Hence, if P be the point (A, k), we have
h = — — L a and k = t zr ^•
A. + 1 \+ 1
Hence for all values of A., i.e. for all positions of the
straight line A'B\ we have
Ji_k
a h '
But this is the condition that the point {h, A), i.e. P,
should lie on the straight line
a h '
The straight line is therefore the required locus.
105. Ex. A variable straight line is drawn through
a given point to cut two fixed straight lines in R and S ;
on it is taken a j)oint P such that
OP''OR'^OS''
shew that the locus of P is a third fixed straight livie.
Take any two fixed straight lines, at right angles and
passing through 0, as the axes and let the equation to the
two given fixed straight lines be
Ax + By + C=0,
and A'x + B'y + G' = 0.
Transforming to polar coordinates these equations are
1 AcosO + B^ine , I A' cos 6 + B' sin 6
-= -^ and - = y^, .
r C r C
THE STRAIGHT LINE. PROBLEMS. 85
1 1
If the angle XOE be 6 the values of ^r^ and ;=r-5 are
therefore
^ cos ^ + ^ sin A' cos $ + B' sin 6
-^ and ~ .
We therefore have
2 _ ^cos^ + .gsin^ A' cos + B' sin 6
0P~~ C C'
fA A'\ a (^ ^'\ ' a
The equation to the locus of P is therefore, on again
transforming to Cartesian coordinates,
_ (A A'\ (B B\
and this is a fixed straight line.
EXAMPLES. XL
The base BG (=:2a) of a triangle ABC is fixed; the axes being
BC and a perpendicular to it through its middle point, find the locus
of the vertex A^ when
1. the difference of the base angles is given ( = a).
2. the product of the tangents of the base angles is given ( = X).
3. the tangent of one base angle is m times the tangent of the
other.
4. m times the square of one side added to n times the square of
the other side is equal to a constant quantity c^.
From a point P perpendiculars TM and PN are drawn upon two
fixed lines which are inclined at an angle w, and which are taken as
the axes of coordinates and meet in ; find the locus of P
5. if Oll^ ON be equal to 2c. 6. if OM- ON be equal to 2d.
7. if PM + PN be equal to 2c. 8. if P^I - P^V be equal to 2c.
9. if MN be equal to 2c.
10. if MN pass through the fixed point (a, b).
11. if MN be parallel to the given line y = mx.
86 COORDINATE GEOMETRY. [Exs.
12. Two fixed points A and B are taken on the axes such that
OA = a and OB = h; two variable points A' and B' are taken on the
same axes; find the locus of the intersection of AB' and A'B
(1) when OA' + OB' = OA + OB,
and (2) when __,__=A__.
13. Through a fixed point P are drawn any two straight lines to
cut one fixed straight Hne OX in A and B and another fixed straight
line OY in. C and D ; prove that the locus of the intersection of the
straight lines AG and BD is a straight line passing through 0.
14. OX and OY are two straight lines at right angles to one
another; on OF is taken a fixed point A and on OX any point B;
on AB an equilateral triangle is described, its vertex C being on the
side of AB away from 0. Shew that the locus of is a straight
line.
15. If a straight line pass through a fixed point, find the locus of
the middle point of the portion of it which is intercepted between two
given straight lines.
16. A and B are two fixed points; if PA and PB intersect a
constant distance 2c from a given straight line, find the locus of P.
17. Through a fixed point are drawn two straight lines at right
angles to meet two fixed straight lines, which are also at right angles,
in the points P and Q. Shew that the locus of the foot of the
perpendicular from on PQ is a straight line.
18. Find the locus of a point at which two given portions of the
same straight line subtend equal angles.
19. Find the locus of a point which moves so that the difference
of its" distances from two fixed straight lines at right angles is equal
to its distance from a fixed straight line.
20. A straight line AB, whose length is c, slides between two
given oblique axes which meet at ; find the locus of the orthocentre
of the triangle OAB.
21. Having given the bases and the sum of the areas of a number
of triangles which have a common vertex, shew that the locus of this
vertex is a straight line.
22. Through a given point a straight line is drawn to cut two
given straight lines in R and S ; find the locus of a point P on this
variable straight line, which is such that
(1) 20P.= 0R+0S,
and (2) 0P^=0R.0S.
XI.] THE STRAIGHT LINE. EXAMPLES. 87
23. Given n straight lines and a fixed point 0; through is
drawn a straight line meeting these lines in the points K^, R^, R.^,
...Bm and on it is taken a point R such that
n 1 1 1 1
OR OR^ OR^ OR^ OR^'
shew that the locus of i2 is a straight line.
24. A variable straight line cuts off from n given concurrent
straight lines intercepts the sum of the reciprocals of which is con-
stant. Shew that it always passes through a fixed point.
25. If a triangle ABC remain always similar to a given triangle,
and if the point A be fixed and the point B always move along a
given straight line, find the locus of the point C.
26. -A. right-angled triangle ABC, having C a right angle, is of
given magnitude, and the angular points A and B slide along two
given perpendicular axes; shew that the locus of G is the pair of
straight lines whose equations are y = ±-x.
27. Two given straight lines meet in 0, and through a given point
P is drawn a straight line to meet them in Q and R; if the
parallelogram OQSR be completed find the equation to the locus
of R.
28. Through a given point is drawn a straight line to meet two
given parallel straight lines in P and Q ; through P and Q are drawn
straight lines in given directions to meet in R ; prove that the locus of
R isa, straight line.
CHAPTER VI.
ON EQUATIONS REPRESENTING TWO OR MORE
STRAIGHT LINES.
106. Suppose we have to trace the locus represented
by the equation
2/2_3a;2/ + 2a;2 = (1).
This equation is equivalent to
{y-x){y-2x) = (2).
It is satisfied by the coordinates of all points which
make the first of these brackets equal to zero, and also by
the coordinates of all points which make the second
bracket zero, i.e. by all the points which satisfy the
equation
2/-^ = (3),
and also by the points which satisfy
y-2x = (4).
But, by Art. 47, the equation (3) represents a straight
line passing through the origin, and so also does equa-
tion (4).
Hence equation (1) represents the two straight lines
which pass through the origin, and are inclined at angles of
45° and tan~^ 2 respectively to the axis of x.
107. Ex. 1. Trace the locus xy = 0. This equation
is satisfied by all the points which satisfy the equation
x — O and by all the points which satisfy y — 0, i. e. by
all the points which lie either on the axis of y or on the
axis of X.
EQUATIONS EEPRESENTING STRAIGHT LINES. 89
The required locus is therefore the two axes of coordi-
nates.
Elx. 2. Trace the locus x^ — hx+ &=^ 0. This equation
is equivalent to {x — 2) (a; — 3) = 0. It is therefore satisfied
by all points which satisfy the equation a; — 2 = and also
by all the points which satisfy the equation a? — 3 = 0.
But these equations represent two straight lines which
are parallel to the axis of y and are at distances 2 and 3
respectively from the origin (Art. 46).
Ex. 3. Trace the locus xy - ix — by ■¥ 'lO = 0. This
equation is equivalent to {x — 5) (2/ — 4) = 0, and therefore
represents a straight line parallel to the axis of y at a
distance 5 and also a straight line parallel to the axis of x
at a distance 4.
108. Let us consider the general equation
ax^ + 2hxy + hy'^ ^ (1).
On multiplying it by a it may be written in the form
(«2x.2 + 2ahxy + hhf) - (A^ - ah) y'^ = 0,
L e. \{ax + hy) + y Jh^ — ab\ {(ax + hy) — y Jh^ — ab\ = 0.
As in the last article the equation (1) therefore repre-
sents the two straight lines whose equations are
ax + hy + y Jh^ — ab = (2),
and ax + hy — y J li^ — ab = (3),
each of which passes through the origin.
For (1) is satisfied by all the points which satisfj'- (2),
and also by all the points which satisfy (3),
These two straight lines are real and different if h^>ab,
real and coincident if h^ = ah, and imaginary if h^<ab.
[For in the latter case the coefiicient of y in each of the
equations (2) and (3) is partly real and partly imaginary.]
In the case when h^ < ab, the straight lines, though
themselves imaginary, intersect in a real point. For the
origin lies on the locus given by (1), since the equation (1)
is always satisfied by the values x = and y = 0.
90 COORDINATE GEOMETRY.
109. An equation such as (1) of the previous article,
which is such that in each term the sum of the indices of x
and y is the same, is called a homogeneous equation. This
equation (1) is of the second degree; for in the first term
the index of cc is 2 ; in the second term the index of both x
and 2/ is 1 and hence their sum is 2 ; whilst in the third
term the index of y is 2.
Similarly the expression
3a^ + ix^y — 5xy'^ + 9y^
is a homogeneous expression of the third degree.
The expression
Scc^ + 4:x'^y — 5xy^ + 9y^ — 7xy
is not however homogeneous ; for in the first four terms
the sum of the indices is 3 in each case, whilst in the last
term this sum is 2.
From Art. 108 it follows that a homogeneous equation
of the second degree represents two straight lines, real and
different, coincident, or imaginary.
110. The axes being rectangular, to jind the angle
between the straight lines given by the equation
aa? + 'ihxy + by- = (1).
Let the separate equations to the two lines be
y—miX = and y~ni^x — (2),
so that (1) must be equivalent to
b (y — mjx) (y — tUox) = (3).
Equating the coefficients of xy and x^ in (1) and (3), we
have
— b {nil + ^2) = 2/i, and bm^n^ = a,
xu i. 2h ^ a
so that m,. + m.y = — — , ana m^m^ — -j .
b b
CONDITIONS OF PERPENDICULARITY. 91
If be the angle between the straight lines (2) we
have, by Art. 66,
tan 6 = ^^ ~ ^^^^ =_- n/(^i + ^2)^ - ^rn-jn ^
1 + m^m^ 1 + m-{nio,
J
F~T 2Vh2-ab
.(4).
1 0^ a + b
Hence the required angle is found.
111. Condition that the straight lines of the 'previous
article may he iV) 'perpendicular, and (2) coincident.
(1) If« + 5=:0 the value of tan ^ is cx) and hence is
90° j the straight lines are therefore perpendicular.
Hence two straight lines, represented by one equation,
are at right angles if the algebraic sum of the coefficients of
^ and 2/^ be zero.
For example, the equations
x^ — if = and Qx^ + llxy — 6^- —
both represent pairs of straight lines at right angles.
Similarly, whatever be the value of h, the equation
a? + Ihx'y — y^ = 0,
represents a pair of straight lines at right angles.
(2) If h^ = ah, the value of tan is zero and hence 6 is
zero. The angle between the straight lines is therefore
zero and, since they both pass through the origin, they are
therefore coincident.
This may be seen directly from the original equation.
For if A^ = ah, i.e. h = J ah, it may be written
ax^ 4- '^Jah x'y + hy- =^0,
i.e. [Jax-\- JhyY=0.
which is two coincident straight lines.
92
COORDINATE GEOMETRY.
112. Tojind the equation to the straight lines bisecting
the angle between the straight lines given by
ax'' + 2hxy + by^ = (1).
Let the equation (1) represent the two straight lines
L^OM^ and LJ)M.^ inclined at angles B^ and 0.^ to the axis
of £c, so that (1) is equivalent to
b{y -X tan 6-^ (y - x tan 0^ = 0.
Hence
tan 6^ + tan ^o = — ;- , and tan 0^ tan 6^ — -. . .(2).
6 1 ^ b '
Let OA and OB be the required bisectors.
Since z AOL^ = / L^OA,
:. iA0X-e^ = 6^~- lAOX.
:. 2 lAOX^-B^ + e,^.
Also z ^(9X = 90° + z .4 OX.
:. 2 z^OX=180° + ^i + ^2.
Hence, if 6 stand for either of the angles AOX or BOX,
we have
n/i J. /A A\ t^i^ ^1 + ■tan ^2 2A
tan 2^ = tan {0^ + ^o) = r-^-a^ — ^ = " I '
' 1 - tan t/j tan t'a 6 — «
by equations (2).
But, if (x, 2/) be the coordinates of any point on either
of the lines OA or OB, we have
tan^=^.
X
TWO STRAIGHT LINES. 93
2h , ^^ 2tan^
tan 2^
a 1 — tan^
a; 2xy
*? *> 9. 5
a?"
x2 - y2 xy
I.e. Z-- = -r— .
a~b h
This, being a relation holding between the coordinates
of any point on either of the bisectors, is, by Art. 42, the
equation to the bisectors.
113. The foregoing equation may also be obtained in the follow-
ing manner :
Let the given equation represent the straight hnes
y-m^x=0 and y-ni^x^O (1),'
so that iJii + m2= - Y' ^-nd m^m^^- (2).
The equations to the bisectors of the angles between the straight
lines (1) are, by Art. 84,
y-ntj^x _ y-m^x _ y-vi^x _ y-m^x
or, expressed in one equation,
j y-vi^x _ y-jn^x \ J y -m^x y-iihx \_^
{y-m^xf {y-m^-_
1 + m-^ l + ma^
i.e. (1 + m^) {y^ - 2m^xy + m^x^) - (1 + m-^) {y^ - ^m^xy + m^x'^) = 0,
i. e. (m^2 - m^^) {pt^ -y^) + 2 (m^Tng - 1) (Wj - m^) xy = 0,
i. e. (mj + W2) {x^ -y^) + 2 (wij W2 - 1) ici/ = 0.
Hence, by (2), the required equation is
-2fe , „ .,. „ /a
6
{x^-y-') + 2{~-ljxy = 0,
x^-y"^ _ xy
I.e. — — ^-— .
a-o h
94 COORDINATE GEOMETRY.
EXAMPLES. XII.
Find what straight lines are represented by the following equations
and determine the angles between them.
1. x^-7xy + 12tf=0. 2. 4a;2-24'cz/+ll?/2 = 0.
3. SSx^--71xij-Uif = 0. 4. x^-&x^ + llx-Q = 0.
5. 2/2-16 = 0. 6. tf-xif-Ux^y + 24:X^==0.
7. x^ + 2xysecd + y'^=0. 8. x'^ + 2xy cotd + 7f = 0.
9. Find the equations of the straight lines bisecting the angles
between the pairs of straight lines given in examples 2, 3, 8, and 9.
10. Shew that the two straight lines
x^ (tan2 + cos2 ^j _ 2xy tan d + y^ sin^ ^ =
make with the axis of x angles such that the difference of their
tangents is 2.
11. Prove that the two straight lines
{x^ + y^) (cos^ d sin2 a + sin^ 6) =■■ {x t&n a - y sin 6)-
include an angle 2a.
12. Prove that the two straight lines
x^ sia^acoa^9 + 4iXy sin a sin d + y'^[4: cos a - (1 + cos a)^ cos2^] =
meet at an angle a.
GENERAL EQUATION OF THE SECOND DEGREE.
114. The most general expression, which contains
terms involving x and y in a degree not higher than the
second, must contain terms involving x^, xy, 7/~, x, y, and a
constant.
The notation vv^hich is in general use for this ex-
pression is
aa^ + 2hxy + hy"^ + 2gx + yy + c (1).
The quantity (1) is known as the general expression of
the second degree, and when equated to zero is called the
general equation of the second degree.
The student may better remember the seemingly
arbitrary coefficients of the terms in the expression (1)
if the reason for their use be given.
GENERAL EQUATION TO TWO STRAIGHT LINES. 95
The most general expression involving terms only of
the second degree in x, y, and z is
aa^ + h'lf' + Gz^ + %fyz + '^gzx + Vixy (2),
where the coefficients occur in the order of the alphabet.
If in this expression we put ;:; equal to unity we get
ao^ -\-hy^ + G + %fy + ^gx + 2hxy,
which, after rearrangement, is the same as (1).
Now in Solid Geometry we use three coordinates x, y,
and z. Also many formulae in Plane Geometry are derived
from those of Solid Geometry by putting z equal to unity.
We therefore, in Plane Geometry, use that notation
corresponding to which we have the standard notation in
Solid Geometry.
115. In general, as will be shewn in Chapter XV.,
the general equation represents a Curve-Locus.
If a certain condition holds between the coefficients of
its terms it will, however, represent a pair of straight lines.
This condition we shall determine in the following
article.
116. To find the condition that the general equation
of the second degree
ax^ + 2hxy + hy'^ + Igx + 2/^/ + c == (1)
Tnay rejyresent tivo straight lines.
If we can break the left-hand members of (1) into two
factors, each of the first degree, then, as in Art. 108, it
will represent two straight lines.
If a be not zero, multiply equation (1) by « and arrange
in powers oi x; it then becomes
a^a? + 2ax {hy + g) = — ahy^ — 2afy — ac.
On completing the square on the left hand we have
«V ^ 2ax {hy + g) + (hy + gf = y- {li? — ah)
■\-2y{gh-af)^g''~-ac,
i.e.
{ax-\-hy+g)=^Jy''{h''-ab) + 2y{gh- of) + g^'-ac ...{2).
96 COORDINATE GEOMETRY.
From (2) we cannot obtain x in terms of y, involving
only terms of the Jirst degree, unless the quantity under the
radical sign be a perfect square.
The condition for this is
i. e. gVi^ — 2qfgh + o?f'^ = g^li? — ahg"^ — acli? + a^bc.
Cancelling and dividing by a, we have the required
condition, viz.
abc + 2fgh - af 2 - bg2 - ch2 = O (3).
117. The foregoing condition may be otherwise obtained thus :
The given equation, multiplied by (a), is
a?x'^-{-2ahxy + aby'^ + 2agx^-2afij + ac = (4).
The terms of the second degree in this equation break up, as in
Art. 108, into the factors
ax + liy-y /Jh^-ab and ax + hy+y fjh^-ab.
If then (4) break into factors it must be equivalent to
{ax + {h-^JhF^ab)y + A}{ax + {h+s/h^-ab)y + B} = 0,
where A and B are given by the relations
a{A+B) = 2ga (5),
A{h+ ^W^^) +B{h- ^h^ - ab) = 2fa (6),
and AB = ac (7).
The equations (5) and (6) give
, , ^ 2fa-2gh
A+B = 2g, and A - B=^^-—=A=..
/^/h^ — ab
The relation (7) then gives
Aac = 4:AB = {A+B)^-{A-By-
~^^ ^ h^-ab '
i.e. {fa-ghf = {g-~ac){h^-ab),
which, as before, reduces to
abc + 2fgh - ap - bg"^ - cli^=^^.
Ex. If a be zero, prove that the general equation will represent
two straight lines if
If both a and 6 be zero, prove that the condition is 2fg - ch~0.
GENERAL EQUATION TO TWO STRAIGHT LINES. 97
118. The relation (3) of Art. 116 is equivalent to the
expression
<x, h, g
This may be easily verified by writing down the value
of the determinant by the rule of Art. 5.
A geometrical meaning to this form of the relation (3)
will be given in a later chapter.
The quantity on the left-hand side of equation (3) is
called the Discriminant of the General Equation.
The general equation therefore represents two straight
lines if its discriminant be zero.
119. Ex. 1. Prove that the equation
12a;2 + Ixy - lOy"^ + 13a; + 45?/ - 35 =
represents tioo straight lines, and find the angle between them.
Here
a=12, /i=|, &=-10, g=^, /=-*/, and c=- 35.
Hence abc + 2fgh - ap - bg^ - ch^
= 12 X ( - 10) X ( - 35) + 2 X ¥- x V- x ^ - 12 x (V)^ - ( - 10) x (V)^
-(-35)(i)2
= 4200 + ^V- - 6075 + i^-^ + i J^5-
= - 1875 + ^^-^^ = 0.
The equation therefore represents two straight lines.
Solving it for x, we have
o,, 7y + 13 f7y + lBY_ 10y^-4.mj + S5 f ly + is y
^ +"^-12" + V ~ 2^-; - — 12 — + \2r~)
_/2%--43\2
~l 24 J •
. , , 7y + 13 _ 2 3y-43
•• ^+-2^-=^-- 24""'
2?/ - 7 - 5w + 5
I. e. x= -— — or — -. .
o 4
The given equation therefore represents the two. straight lines
3x=2y-7 and 4x= -5y + 5.
L. r
98 COORDINATE GEOMETRY.
The "m's" of these two lines are therefore f and -f , and the
angle between them, by Art. 66,
Ex. 2. Find the value of h so that the equation
6a;2 + 2hxy + 12y^ + 22a; + 31?/ + 20 =
may represent two straight lines.
Here
a = 6, i = 12, ^ = 11, /=V-. andc = 20.
The condition (3) of Art. 116 then gives
20;i2- 341/1 +H*'-^=o,
i.e. (/i -V^) (20/1- 171) = 0.
Hence 7i= V- or ^.
Taking the first of these values, the given equation becomes
6^2 + nxy + 12y^ + 22x + 31?/ + 20 = 0,
i.e. (2a; + 3?/ + 4) (3a; + 4?/ + 5) = 0.
Taking the second value, the equation is
20a;2 + 57a;?/ + 40?/2 + ^fa a; + H-2/ + -t- = 0,
i.e. (4a; + o?/ + -V)(5a; + 8?/ + 10) = 0.
EXAMPLES. XIII.
Prove that the following equations represent two straight lines ;
find also their point of intersection and the angle between them.
1. 6?/2-rci/-a;2 + 30?/ + 36 = 0. 2. a;2 - 5;r?/ + 4?/2 + a; + 2?/ - 2 = 0.
3. 3?/2-8a;?/-3a;2-29a; + 3?/-18 = 0.
4. y^ + ^y - 2a;2 - 5x-y -2 = 0.
5. Prove that the equation
a;2 + 6a;?/ + 9?/2 + 4a; + 12?/ - 5 =
represents two parallel lines.
Find the value of k so that the following equations may represent
pairs of straight lines :
6. 6a;2+lla;?/-10?/2 + a; + 31?/ + ^ = 0.
7. 12a;2-10a;?/ + 2?/2 + lla;-5?/ + A; = 0.
8. 12a;2 + A;a;?/ + 2?/2 + lla;-5?/ + 2:=0.
9. 6a;2 + a;?/ + A-?/2-ll.r + 43?/-35 = 0.
[EXS. XIII.] EXAMPLES. 99
10. kxy-8x + 9y-12 = 0.
11. x'^ + is^-xij + 'if-5x-7y + k=:0.
12. 12x^ + xy-Qif-2dx + 8y + k = 0.
13. 2x^ + oiyy-y'^ + kx + 6y-d = 0.
14. x^ + kxy + y^-5x-ly + Q = 0.
15. Prove that the equations to the straight lines passing through
the origin which make an angle a with the straight line y + x = are
given by the equation
x^ + 2xy sec 2a + y^ = Q.
16. What relations must hold between the coordinates of the
equations
(i) ax^ + hy'^ ■^cx-\-cy = 0,
and (ii) ay'^ + lxy + dy -{-ex=zQ^
so that each of them may represent a pair of straight lines ?
17. The equations to a pair of opposite sides of a parallelogram
are
a;2-7a; + 6 = and 2/^- 14?/ +40 = ;
find the equations to its diagonals.
120. To prove that a homogeneous equation of the nth
degree represents n straight li7ies, real or imaginary, which
all pass through the orighi.
Let the equation be
y'^ + A^xy''-^ + A^xY'-^ + A.^x'y''-^ 4- . . . + A.^x'' =-- 0.
On division by x^, it may be written
This is an equation of the nth degree in - , and hence
JO
must have n roots.
Let these roots be -m^, m^, mg, ... m„. Then (C. Smith's
Algebra, Art. 89) the equation (1) must be equivalent to
the equation
t
The equation (2) is satisfied by all the points which
satisfy the separate equations
'^)6-"'^)(l-™0- (!-'"«)=•'■■•<'>•
^ _ mi = 0, ^ - m. = 0, . . . ^ - m,, - 0,
/v. ^ 'T* « 5
tLf *^ tKf
7—2
100 COORDINATE GEOMETRY.
i.e. by all the points which lie on the n straight lines
y — nijX = 0, y — m^x = 0, ... y — in^x = 0,
all of which pass through the origin. Conversely, the
coordinates of all the points which satisfy these n equa-
tions satisfy equation (1). Hence the proposition.
121. Ex. 1. The equation
which is equivalent to
{y-x){y-'lx){y-^x) = Q,
represents the three straight lines
y-x=.^, ?/-2a: = 0, and y-3a; = 0,
all of which pass through the origin.
Ex. 2. The equation y^ - 5y^ + 6y = 0,
i.e. y{y-2){y-S) = 0,
similarly represents the three straight lines
y=0, y = 2, and y = S,
all of which are parallel to the axis of x.
122. To find the equation to the two straight lines
joining the origin to the points in which the straight line
Ix + my = n (1)
meets the locus whose equation is
ax^ + 2hxy + hy"^ + "Igx + 2fy + c = (2).
The equation (1) may be written
Ix + my
n
= 1 (3).
The coordinates of the points in which the straight line
meets the locus satisfy both equation (2) and equation (3),
and hence satisfy the equation
ax' + Vum) + by^ + 2 {gx +/y) '-^^ + c ('-^^)' =
......(4).
[For at the points where (3) and (4) are true it is clear
that (2) is true.]
Hence (4) represents so7ne locus which passes through
the intersections of (2) and (3).
STRAIGHT LINES THROUGH THE ORIGIN. 101
But, since the equation (4) is homogeneous and of the
second degree, it represents two straight lines passing
through the origin (Art. 108).
It therefore must represent the two straight lines join-
ing the origin to the intersections of (2) and (3).
123. The preceding article may be illustrated geo-
metrically if we assume that the equation (2) represents
some such curve as PQRS in the figure.
■"'/O
Let the given straight line cut the curve in the points
P and Q.
The equation (2) holds for all points on the curve PQRS.
The equation (3) holds for all points on the line PQ.
Both equations are therefore true at the points of
intersection P and Q.
The equation (4), which is derived from (2) and (3),
holds therefore at P and Q.
But the equation (4) represents two straight lines, each
of which passes through the point 0.
It must therefore represent the two straight lines OP
and OQ.
124. Ex. Prove that the straight lines joining the origin to the
points of intersection of the straight line x-y=:2 and the curve
5a;2 + 12a;i/ - 82/2 + 8a; - 4y + 12 =
make equal angles with the axes.
As in Art. 122 the equation to the required straight lines is
5a;2 + 12a;?/-8i/2+(8a;-4i/)^^ + 12fc^Y = (1),
102 • COORDINATE GEOMETRY.
For this equation is homogeneous and therefore represents two
straight lines through the origin; also it is satisfied at the points
where the two given equations are satisfied.
Now (1) is, on reduction,
so that the equations to the two lines are
y = 2x and y= -2x.
These lines are equally inclined to the axes.
125. It was stated in Art. 115 that, in general^ an
equation of the second degree represents a curve- line,
including (Art. 116) as a particular case two straight lines.
In some cases however it will be found that such
equations only represent isolated points. Some examples
are appended.
EjX. 1. What is represented hy the locus
{x-y + cf+{x + y-cY = 01 (1).
We know that the sum of the squares of two real
quantities cannot be zero unless each of the squares is
separately zero.
The only real points that satisfy the equation (1)
therefore satisfy both of the equations
cc — 2/ + c = and x + y—G = 0.
But the only solution of these two equations is
re = Oj and y = c.
The only real point represented by equation (1) is therefore
(0,c).
The same result may be obtained in a different manner.
The equation (1) gives
{x-y-VGY^-(x + y-cf,
i.e. x — y + c = ^ V— 1 {x + y — c).
It therefore represents the two imaginary straight lines
x{l- J'^) -y(l + J^) + c (1 + J^) = 0,
and X (1 + J^l)-y (I - J^) + c(l - J~l) = 0.
EQUATIONS REPRESENTING ISOLATED POINTS. 103
Each of these two straight lines passes through the
real point (0, c). We may therefore say that (1) represents
two imaginary straight lines passing through the point
(0, 0).
£jX. 2. What is represented hy the equation
As in the last example, the only real points on the locus
are those that satisfy both of the equations
oc^ — a^ — and y^ — h^ = 0,
i.e. x = =i= a, and y = d=h.
The points represented are therefore
(a, h), {a, —h), (—a, b), and (—a, —6).
Ex. 3. WTiat is represented by the equation
The only real points on the locus are those that satisfy
all three of the equations
x = 0, 2/=0, and a = 0.
Hence, unless a vanishes, there are no such points, and
the given equation represents nothing real.
The equation may be written
a? + y^ = — a^y
so that it represents points whose distance from the origin
is asl—\. It therefore represents the imaginary circle
whose radius is asJ—1 and whose centre is the origin.
126. Es. 1. Obtain the condition that one of the straight lines
given hy the equation
ax^-\-2hxy + by^ = (1)
may coincide with one of those given by the equation
a'x^ + 2h'xy + bY=0 (2).
Let the equation to the common straight line be
y-m^x = (3).
The quantity y -m^x must therefore be a factor of the left-hand of
both (1) and (2), and therefore the value y = m^x must satisfy both (1)
and (2).
104 COORDINATE GEOMETRY.
"We therefore have
bmi^ + 2hm^ + a=:0 (4),
and b'm^^ + 2h'm^ + a' = (5).
Solving (4) and (5), we have
wij^ _ '^h _ ^
2 {ha' - h'a) ~ aV -a'b~2 {bh' - b'h) '
^ ha'-h'a _ ^_ f ab'-a'b ]^
■*• bh' -b'h~^~\2 {bh' - b'h)( '
so that we must have
{ab' - a'&)2 = 4 {ha' - h'a) {bh' - b'h) .
Ex. 2. Prove that the equation
m{x^-Sxy^) + y^-3x^y=0
represents three straight lines equally inclined to one another.
Transforming to polar coordinates (Art. 35) the equation gives
m {GQS^d- 3 cos d sin2^) + sin3^ - 3 cos^^ sin ^ = 0,
i.e. m(l-3tan2^) + tan3^-3tan^ = 0,
3tan^-tan3^ ^ „^
If m=tan a, this equation gives
tan 3^ = tan a,
the solutions of which are
3^ = a, or 180° + a, or 360° + a,
i.e. ^ = |, or 60° + ^, or 120° + ^.
The locus is therefore three straight lines through the origin
inclined at angles
^, 60° + |, and 120° + ^
to the axis of x.
They are therefore equally inclined to one another.
Ex. 3. Prove that two of the straight lines represented by the
equation
ax'^ + bx^y + cxy^ + dy^ = (1)
will be at right angles if
a^ + ac + hd + d^ = 0.
Let the separate equations to the three lines be
y-miX = 0, y-m2X=0, and y-m^x=0,
EXAMPLES. 105
so that the equation (1) must be equivalent to
d{y - m-^x) (y - m^) (y - m^x) = 0,
c
and therefore mj^+m2+m^= -- (2),
Wom3 + m3?7ij + 77ijm2 = -^ (3)i
and m^m^m^ = - -^ (4).
If the first two of these straight lines be at right angles we have,
in addition,
711^111.2= -1 (5).
From (4) and (5), we have
a
and therefore, from (2),
c a c + a
The equation (3) then becomes
a f c + a\ _b
i.e. a^ + ac + bd + d^ = 0.
EXAMPLES. XIV.
1. Prove that the equation
y^-x^ + 3xy (y -x) =
represents three straight lines equally inclined to one another.
2. Prove that the equation
y^ (cos a + fJ3 sin a) cos a-xy (sin 2a - ^^3 cos 2a)
+ x^ (sin a - ^3 cos a) sin a =
represents two straight lines inclined at 60° to each other.
Prove also that the area of the triangle formed with them by the
straight line
(cosa-/sy3 sin a) 2/ -(sin a + ;^3cos a)a; + a-=0
a2
V3'
and that this triangle is equilateral.
3. Shew that the straight lines
(^2 _ 3^2) ^2 + sABxy + [B'^ - 3A^) y^=0
form with the Hne Ax + By + C = an equilateral triangle whose area
^^ J3{A^ + B^)'
106 COORDINATE GEOMETRY. [ExS.
_ 4. Find the equation to the pair of straight lines joining the
origin to the intersections of the straight line y = mx + c and the curve
Prove that they are at right angles if
2c2 = a2(l + m2).
5. Prove that the straight lines joining the origin to the points
of intersection of the straight line
kx + hy = 2hk
with the curve {^-h)^+{y- k)^ = c^
are at right angles if h^+k^=c^.
6. Prove that the angle between the straight lines joining the
origin to the intersection of the straight line y = 3x + 2 with the curve
x'^ + 2xy + Sy^ + 'ix+8y-ll = istan-i?^.
3
7. Shew that the straight lines joining the origin to the other two
points of intersection of the curves whose equations are
ax^ + 2hxy + by^ + 2gx=0
and a'x'^ + 2h'xy + hY + ^g'x =
will be at right angles if
g{a' + b')-g'{a + b) = 0.
What loci are represented by the equations
8. x^-y^=0. 9. x'^-xy = 0. 10. xy-ay = 0.
11. x^-x^-x + l = 0. 12. x^-xy^ = 0. 13. x^ + y^ = 0.
14. x^ + y^=0. 15. x^y = 0. 16. {x'--l){y^-^)=0.
17. {x^-lf + {y^-4y=0. 18. {y-mx-cY + {y-m'x-c')^=0,
19. {a;2-a3)2(^2_52)2 + c'*(?/2-a2) = 0. 20. {x-a)^-y^=0.
21, (x + y)^-c^=0. 22. r=a sec (<?- a).
23. Shew tliat the equation
bx^-2hxy + ay^=0
represents a pair of straight lines which are at right angles to the pair
given by the equation
ax^ + 2 Jixy + by^ = 0.
24. If pairs of straight lines
x^-2pxy -y^=0 and x^-2qxy-y^=0
be such that each pair bisects the angles between the other pair, prove
thatjpgr= -1.
25. Prove that the pair of lines
a^x'^ + 2h{a + b)xy + b^y^=zO
is equally inclined to the pair
ax^ + 2hxy + by^=0.
XIV.] EXAMPLES. 107
26. Shew also that the pair
is equally inclined to the same pair.
27. If one of the straight lines given by the equation
ax^ + Ihxy + hy^ =
coincide with one of those given by
a'x^ + 2h'xy + bY=0,
and the other lines represented by them be perpendicular, prove that
ha'b' h'ah , — -
-a b-a ^
28. Prove that the equation to the bisectors of the angle between
the straight lines ax^ + 2hxy + by'^=0 is
h {x^ - y^) + {b - a) xy = {ax^ - by^) cos w,
the axes being inclined at an angle w.
29. Prove that the straight lines
ax^-\-1'kxy + by'^=zQ
make equal angles with the axis of a; if 7t = acosw, the axes being
inclined at an angle w.
30. If the axes be inclined at an angle w, shew that the equation
x^ + 2ocy cos w + 2/^cos 2a; =
represents a pair of perpendicular straight lines.
31. Shew that the equation
cos 3a {x^ - dxy^) + sin 3a (y^ - Sx'^y) + 3a (x^ + y^) -^a^ =
represents three straight lines forming an equilateral triangle.
Prove also that its area is 3 sJSa^.
32. Prove that the general equation
ax^ + 2Jixy + by^ + 2gx + 2fy + c=zO
represents two parallel straight lines if
h^ = ab and bg^=^af^.
Prove also that the distance between them is
/ 9'-
V a(a-
ac
{a + b)'
33. If the equation
ax'^ + 2hxy + by^ + 2gx + 2fy + c =
represent a pair of straight lines, prove that the equation to the third
pair of straight lines passing through the points where these meet the
axis is
ax^ - 2hxy + by^ + 2gx + 2fy + c + -^^xy = 0.
108 COORDINATE GEOMETRY. [Exs. XIV.]
34. If the equation
as(^ + 2hxy + by^ + 2gx + 2fy + c =
represent two straight lines, prove that the square of the distance of
their point of intersection from the origin is
35. Shew that the orthocentre of the triangle formed by the
straight lines
ax^ + 2kxy + by^=0 and lx + viy = l
is a point (x\ y') such that
^ _y' _ a + i>
I ~ m~ aw? - 2hlm + bP '
36. Hence find the locus of the orthocentre of a triangle of which
two sides are given in position and whose third side goes through a
fixed point.
37. Shew that the distance between the points of intersection of
the straight Hne
X cos a + y sin a-p =
with the straight lines ax^ + 2 hxy + dy^=0
2pjh^-ab
& cos^a - 2/i cos a sin a + a sin^ a '
Deduce the area of the triangle formed by them.
38. Prove that the product of the perpendiculars let fall from the
point {x', y') upon the pair of straight lines
ax^ + 2hxy + by"^ —
ax''^ + 21tx'y' + by'^
^^ J{a-bf + 4:h^ '
39. Shew that two of the straight lines represented by the
equation
ay"^ + bocy^ + cx^y^ -}-dx^y + ex'^ =
will be at right angles if
(& + d) {ad + be) + {e~ a)^ {a + c + e) = 0.
40. Prove that two of the lines represented by the equation
ax^ + bx^ y + cx^ y^ + dxy^ + ay^ =
will bisect the angles between the other two if
c + Qa=0 and b + d = 0.
41. Prove that one of the lines represented by the equation
ax^ + bx"^ y + cxy^ + dy^ =
will bisect the angle between the other two if
{3a + c)^{bc + 2cd - Sad) = (6+ Sd)^{bc + 2ab - 3ad).
CHAPTER yil.
TRANSFORMATION OF COORDINATES.
127. It is sometimes found desirable in the discussion
of problems to alter the origin and axes of coordinates,
either by altering the origin without alteration of the
direction of the axes, or by altering the directions of the
axes and keeping the origin unchanged, or by altering the
origin and also the directions of the axes. The latter case
is merely a combination of the first two. Either of these
processes is called a transformation of coordinates.
We proceed to establish the fundamental formulae for
such transformation of coordinates.
128. To alter the origin of coordinates without altering
the directions of the axes.
Let OX and Z be the original axes and let the new
axes, parallel to the original, be
O'X' and O'Y'.
Let the coordinates of the new
origin 0\ referred to the original
axes be h and k, so that, if O'L be
perpendicular to OX, we have
OL = h and LO' = h.
Let P be any point in the plane
of the paper, and let its coordinates, referred to the original
axes, be x and y, and referred to the new axes let them be
x' and y'.
Draw PN perpendicular to OX to meet OX' in N'.
Y'
p
N'
X'
N
110
COORDINATE GEOMETRY.
Then
ON^x, NF = y, 0'N' = x, and N'F^y'.
We therefore have
X ^ 0N= OL + O'N' = h + x\
and y = FP = LO' + N'P = k + y'.
The origin is therefore transferred to the point Qi, k) when
we substitute for the coordinates x and y the quantities
X + h and y' + k.
The above article is true whether the axes be oblique
or rectangular.
129. To change the direction of the axes of coordinates,
without changing the origin, both systems of coordinates being
rectangular.
Let OX and OF be the original system of axes and OX'
and OY' the new system, and let
the angle, XOX' , through which
the axes are turned be called 0. Y'
Take any point F in the plane
of the paper.
Draw FN and FN' perpen-
dicular to OX and OX', and also
N'L and N'M. perpendicular to OX and FN.
If the coordinates of F, referred to the original axes,
be X and y, and, referred to the new axes, be xi and y', we
have
ON^x, NF = y, ON'^x', and N'F = y.
The angle
MFN' - 90° - z MN'F^ l MN'O = z XOX' = 6.
We then have
X = 0N= OL -MN'= ON' cobO-N'F sine
= x' cos — y sin (1),
and y = NF = LN' + MF= ON' sin 6 + N'F cos $
= oj' sin ^ + 2/' cos ^ (2).
CHANGE OF AXES. Ill
If therefore in any equation we wish to turn the axes,
being rectangular, through an angle we must substitute
X' cos ^ — y' sin and x' sin ^ + y' cos
for X and y.
"When we have both to change the origin, and also the
direction of the axes, the transformation is clearly obtained
by combining the results of the previous articles.
If the origin is to be transformed to the point (Ji, k)
and the axes to be turned through an angle 6, we have to
substitute
h + X cos — y sin 6 and k + x sin + y cos 6
for X and y respectively.
The student, who is acquainted with the theory of projection of
straight lines, will see that equations (1) and (2) express the fact that
the projections of OP on OX and OY are respectively equal to the
sum of the projections of ON' and N'P on the same two lines.
130. Ex. 1. Transform to parallel axes through the 'point ( - 2, 3)
the equation
2a;2 + 4a;2/ + 5^/2 - 4^ - 22?/ + 7 =: 0.
We substitute x=x' -2 and y=y' + S, and the equation becomes
2 (x' - 2)3 + 4 (a;' - 2) (?/' + 3) + 5 (?/' + 3)2 - 4 (x' - 2) - 22 (i/' + 3) + 7 = 0,
i.e. 2x'^ + 4xy + 5y'^-22 = 0.
Ex. 2. Transform to axes inclined at 30° to the original axes the
equation
x'^ + 2fj3xy-y'^=2a^.
For X and y we have to substitute
flj' cos 30° - ?/ sin 30° and a;' sin 30° + ?/' cos 30°,
t.e. — ^^r — - and ^--^— .
The equation then becomes
(a:V3 -2/T + 2 V3 (^V3 -2/') (^' + 2/V3) - (^' + 2/V3)'=8a2,
i.e, x'^-y'^ = a^.
112 COORDINATE GEOMETRY.
EXAMPLES. XV.
1. Transform to parallel axes through the point (1, -2) the
equations
(1) y'^-4:X + 4y + 8 = 0,
and (2) 2x^ + y^-4:X + 4y = 0.
2. What does the equation
{x-a)^+{y-b)^=c^
become when it is transferred to parallel axes through
(1) the point {a-c, b),
(2) the point {a, b-c)?
3. What does the equation
{a-b){x^ + y^-)-2abx=0
become if the origin be moved to the point ( , ) ?
4. Transform to axes inclined at 45° to the original axes the
equations
(1) x^-y^ = a\
(2) nx^-l&xy + ny^ = 225,
and (3) y^ + x^ + 6x^y^=2.
5. Transform to axes inclined at an angle a to the original axes
the equations
(1) x^+y^=r^, .
and (2) a^ + 2xy tan 2a- y^=aK
6. If the axes be turned through an angle tan~i 2, what does the
equation Axy - 3x^ = a^ become ?
7. By transforming to parallel axes through a properly chosen
point {h, k), prove that the equation
12ciP-10xy + 2y^ + llx-5y + 2 = 0-
can be reduced to one containing only terms of the second degree.
8. Find the angle through which the axes may be turned so that
the equation Ax + By + 0=0
may be reduced to the form a; = constant, and determine the value of
this constant.
131. The general proposition, which is given in the
next article, on the transformation from one set of oblique
axes to any other set of oblique axes is of very little
importance and is hardly ever required.
CHANGE OF AXES. 113
■*132. To change from one set of axes, inclined at an
angle w, to another set, inclined at an angle w', the origin
remaining unaltered.
\ __ — \ /""1M
O M N L X
Let OX and (9Fbe the original axes, OX' and OY' the
new axes, and let the angle XOX' be 0.
Take any point P in the plane of the paper.
Draw PN and PN' parallel to 07 and OY' to meet OX
and OX' respectively in N and iV", PL perpendicular to OX,
and N'M and X'M' perpendicular to OL and LP.
Now
z PNL = L YOX = 0), and PN'M' = Y'OX = oi' + B.
Hence if
OX^x, NP^y, ON'^x, s.rvAN'P^y',
we have y s>\n.oi = NP ^irna^ LP = MX' + M'P
= OX' sin + X'P sin (w' + 6),
so that y sin w = cc' sin 6 + y' sin (w' + ^) (1).
Also
x-ryco&oi^OX+XL=^OL=^OM-\-X'M'
= x cosO + y'cos{oy' + e) (2).
Multiplying (2) by sinw, (1) by cosw, and subtracting,
we have
X sin (o = x sin (w — ^) + y' sin (oi — w —6) (3).
[This equation (3) may also be obtained by drawing a perpen-
dicular from P upon OT and proceeding as for equation (1).]
The equations (1) and (3) give the proper substitutions
for the change of axes in the general case.
As in Art. 130 the equations (1) and (2) may be obtained by
equating the projections of OP and of ON' and N'P on OX and a
straight line perpendicular to OX.
L. , 8
114 COORDINATE GEOMETKY.
'^133. Particular' cases of the preceding article.
(1) Suppose we wish to transfer our axes from a
rectangular pair to one inclined at an angle w'. In this
case (0 is 90°, and the formulse of the preceding article
become
x = x' cos 6 + y' cos (w' + 6),
and y = ^' sin 6 + y sin (w' + 6).
(2) Suppose the transference is to be from oblique
axes, inclined at w, to rectangular axes. In this case co' is
90°, and our formulse become
X sin isi = X sin (w — ^) — y' cos (w — 0)^
and y sin o> = a:;' sin 6 + y' cos ^.
These particular formulse may easily be proved in-
dependently, by drawing the corresponding figures.
Ex. Transform the equation -2 - f^ = l from rectangular axes to
axes inclined at an angle 2a, the new axis of x being inclined at an angle
— a to the old axes and sin a being equal to — , .
Here 6= -a and w' = 2a, so that the formulse of transformation
(1) become
a; = (a;' + y) cos a and y = [y' - x') sin a.
Since sin a = , , we have cos a = , , and hence the
x/a2+&2 Ja'^ + b'^
given equation becomes
i.e. x'y'=i{a'^ + b^).
■*134. The degree of an equation is unchanged hy any
transformation of coordinates.
For the most general form of transformation is found
by combining together Arts. 128 and 132, Hence the
most general formulse of transformation are
, sin (oi — 6) , sin (o> — w' — 6)
x = h-¥x ^ + y ^>-7 ,
sm 0) sm w
- _ , sin , sin (w' + &)
and y = k-vx -. — + y ^ .
smo) sinw
CHANGE OF AXES. 115
For X and y we have therefore to substitute expressions
in X and y' of the first degree, so that by this substitution
the degree of the equation cannot be raised.
Neither can, by this substitution, the degree be lowered.
For, if it could, then, by transforming back again, the
degree would be raised and this we have just shewn to be
impossible.
*135. If by any change of axes, without change of origin, the
quantity ax^ + Ihxy + &i/^ become
a'x'^ + 2h'xY + by%
the axes in each case being rectangular, to prove that
a + b = a' + b', and ab-h^ = a'b' -h'^.
By Art. 129, the new axis of x being inclined at an angle 6 to the
old axis, we have to substitute
a;'cos^-2/'sin^ and x' sin + y' cos. 6
for X and y respectively.
Hence ax"^ + 2hxy + by'^
= a{x'cosd-y' sin ef + 2h {x' cose~y' sin d) {x' sin d + y' cos d)
+ b{x'smd + y'cos6)^
= x'^ [a cos2 e + 2h cos Osind + b sin^ d]
+ 2x'y' [-acosdsind + h (cos^ d - sin^i?) + & cos ^ sin 9]
+ y'^ \a sin2 d-2h cos ^ sin ^ + & cos^ d].
We then have
tt' = a cos^ ^ + 2/i cos ^ sin ^ + 6 sin^ (9
= 1 [(a + &) + (« -6) cos 2^ + 2/1 sin 2^] (1),
b' = aBin^d- 2h cos ^ sin ^ + 6 cos^^
= i[(a + 6)-(a-&)cos2^-27isin2^] (2),
and /i'= -acos^sin^ + /i(cos2^-sin2^) + &cos^sin^
= i[2;icos2^-(a-&)sin2^j (3).
By adding ( 1) and (2) , we have «' + &'=« + &.
Also, by multiplying them, we have
4a'&' = (a + 6)2 - { (a - h) cos 2d +-2h sin 26}^.
Hence 4a'&' - 47i'2
= (a + 6)2 _ i{2h sin 2d + {a- b) cos 20]^+ {2h cos 20 -{a- b) sin 20}^
= (a + &)2 _ [(a - 6)2 + 4/i2] = 4a6 - 4h%
so that a'b'-h'^ = ab-h^.
136. To find the angle through which the axes must be turned so
that the expression ax^ + 2hxy + by^ may become an expression in which
there is no term involving x'y'.
8—2
116 COORDINATE GEOMETRY.
Assuming the work of the previous article the coefficient of x'y'
vanishes if h' be zero, or, from equation (3), if
27icos2^ = (a-Z>)sin2^,
2h
i.e. if tan 2(9= -.
a~h
The required angle is therefore
i tan~i
■■(S)
*137. The proposition of Art. 135 is a particular
case, when the axes are rectangular, of the folloM^ing more
general proposition.
If hy any change of axes, without change of origin, the
quantity ax- + ^hxy + 6?/" becomes a'x'^ + 2h'xy + 6'^/^, then
a + b — 2h cos w a' + b' — 2h' cos w'
sin^ lo sin^ w' '
ab-h^ db'-K''
and . o =^ — . o / ,
sm" (i) sm^ o)
0) a7id oi' being the angles between the original and final indrs
of axes.
Let the coordinates of any point P, referred to the
original axes, be x and y and, referred to the final axes, let
them be x and y .
By Art. 20 the square of the distance between P and
the origin is o^ + 2xy cos a> + 2/^, referred to the original axes,
and x'^ + 2xy cos w' + y"^^ referred to the final axes.
We therefore always have
X- + Ixy cos w + 2/^ = a;'^ + ^xy cos m -vy''^ (1).
Also, by supposition, we have
ax"" + 2hxy + by"" = ax'- + 2h'xy' + b'lj^ (2).
Multiplying (1) by \ and adding it to (2), we therefore have
x^ (« + X) + 2xy (h + X cos to) +y^ (b + X)
= x"" (a' + X) + 2x'y' {h' + X cos o>') + y" {b' + X) . . .(3).
If then any value of X makes the left-hand side of (3) a
perfect square, the same value must make the right-hand
side also a perfect square.
But the values of X which make the left-hand a perfect
square are given by the condition
(h + X cos (o)2 -(a + X) {b + X),
EXAMPLES. 117
i.e. by
\2 (1 - cos^ w) + X{a + h-2h cos oi) + ab- h^ = 0,
^„ . a + h — 2hcos(ji ab — h^^ ,,,
i.e. by X^ + X r^ + -r-^ — = (4).
In a similar manner the values of A, which make the
right-hand side of (3) a perfect square are given by the
equation
^a' + b'-2h'cosoy' a'b'-h'^ ^
X' + X r^-^ + . ^ , =0 (5).
sm^ to sm'^ci) ^ '
Since the values of \ given by equation (4) are the same
as the values of X given by (5), the two equations (4) and
(5) must be the same.
Hence we have
a + b — 2h cos cu a' + b' — 2h' cos co'
and
sin^ (o sin^ co'
ab-h2 a'b'-h'2
sin2 0) sin2 cd'
EXAMPLES. XVI.
1. The equation to a straight line referred to axes inclined at 30°
to one another is y = 2x + l. Find its equation referred to axes
inclined at 45°, the origin and axis of x being unchanged.
2. Transform the equation 2x'^ + 3 sjSxy + Sy^ = 2 from axes
inclined at 30° to rectangular axes, the axis of x remaining
unchanged.
3. Transform the equation x^ + xy + y^ = 8 from axes inclined at
60° to axes bisecting the angles between the original axes.
4. Transform the equation y'^-\-4cy cot a - 4cc=0 from rectangular
axes to oblique axes meeting at an angle a, the axis of x being kept
the same.
5. If aj and y be the coordinates of a point referred to a system of
obUque axes, and x' and y' be its coordinates referred to another
system of oblique axes with the same origin, and if the formulae of
transformation be
x=mx' + ny' and y — m'x'-\-n'y\
., . m^ + m'^-1 mm'
prove that » , ,. ., = — 7- .
CHAPTER VIII.
THE CIRCLE.
138. Def. A circle is the locus of a point which
moves so that its distance from a fixed point, called the
centre, is equal to a given distance,
called the radius of the circle.
The given distance is
139. To find the equation to a circle, the axes of coordi-
nates being two straight lines through its centre at right
angles.
Let be the centre of the circle and let a be its radius.
Let OX and OF be the axes of
coordinates.
Let P be any point on the circum-
ference of the circle, and let its coordi-
nates be X and y.
Draw PJf perpendicular to OX and
join OP.
Then (Euc. I. 47)
OM^ + MP' = a',
i.e. x2 + y2 = a2.
This being the relation which holds between the coordi-
nates of any point on the circumference is, by Art. 42, the
required equation.
140. To find the equation to a circle referred to any
rectangular axes.
THE CIRCLE.
119
Let OX and (9 F be the two rectangular axes.
Let C be the centre of the
circle and a its radius.
Take any point P on the
circumference and draw per-
pendiculars CM and FN upon
OX ; let F be the point (x, y).
Draw GL perpendicular to
NF.
Let the coordinates of G be
h and h ; these are supposed to be known.
We have GL = MN= ON- OM=x- h,
and LF = NF-NL = NF-MG^y-h.
Hence, since GL^ + LF^ = GF\
we have (x-h)2+ (y-k)2 = a2 (1).
This is the required equation.
Ex. The equation to the circle, whose centre is the point ( - 3, 4)
and whose radius is 7, is
i.e. a;2 + 2/2 + 6a:-8?/ = 24.
141. Some particular cases of the preceding article may be
noticed :
(a) Let the origin be on the circle so that, in this case,
i.e. h^ + Jc'^=a^.
The equation (1) then becomes
i.e. x'^ + y^-2hx-2ky = 0.
(jS) Let the origin be not on the curve, but let the centre lie on
the axis of x. In this case k = 0, and the equation becomes
{x - lif ■\-y^ = a^.
(7) Let the origin be on the curve and let the axis of re be a
diameter. We now have fe = and a = h, so that the equation becomes
x^ + t/-2hx = 0.
(8) By taking at C, and thus making both h and k Tiero, we
have the case of Art. 139.
120 COORDINATE GEOMETRY.
(e) The circle will touch the axis of x if MG be equal to the
radius, i.e. if k = a.
The equation to a circle touching the axis of x is therefore
x'^ + 7f-2hx-2ky + h'^ = Q.
Similarly, one touching the axis of y is
x^ + y^-2hx-2ky + k^ = 0.
142. To prove that the equation
a^ + 2/'+2^a;+2/2/ + c = (1),
always represents a circle for all values of g,f and c, and to
find its centre and radius. [The axes are assumed to be
rectangular.]
This equation may be written
{a? + 2gx + cf) + {f + 2/2/ +/^) = cf +f' - c,
i.e. {X + gf 4- {y +ff = {JfTp^cf.
Comparing this with the equation (1) of Art. 140, we
see that the equations are the same if
h = -g, h^-f and a = Jg"" -hf^-c.
Hence (1) represents a circle whose centre is the point
(— ^, —f), and whose radius is Jg^-^f^ — c.
If g^ +f' > c, the radius of this circle is real.
If g^ +f'^ = c, the radius vanishes, i. e. the circle becomes
a point coinciding with the point (— ^, —f). Such a circle
is called a point-circle.
If g^ +f^ < c, the radius of the circle is imaginary. In
this case the equation does not represent any real geo-
metrical locus. It is better not to say that the circle does
not exist, but to say that it is a circle with a real centre
and an imaginary radius.
Ex. 1. The equation x"^ + ij^ + 4^x - 6y = can be written in the
form
{x + 2f + iy-Bf = lS = {JlS)\
and therefore represents a circle whose centre is the point ( - 2, 3) and
whose radius is >,yi3.
GENERAL EQUATION TO A CIRCLE. 121
Ex. 2. The equation 45a;2 + 45?/2 - 60x- + SQy + 19 = is equivalent
to
i.e. («'-|)^ + (y + F=f + ^*5-H=/A,
and therefore represents a circle whose centre is the point (f , -|) and
whose radius is —=- .
15
143. Condition that the general equation of the second
degree may represent a circle.
The equation (1) of the preceding article, multiplied by
any arbitrary constant, is a particular case of the general
equation of the second degree (Art. 114) in which there is
no term containing xy and in which the coefficients of x^
and y^ are equal.
The general equation of the second degree in rectangular
coordinates therefore represents a circle if the coefficients
of x^ and y^ be the same and if the coefficient of xy
be zero.
144. The equation (1) of Art. 142 is called the
general equation of a circle^ since it can, by a proper
choice of g, f and c, be made to represent any circle.
The three constants g, f, and c in the general equation
correspond to the geometrical fact that a circle can be found
to satisfy three independent geometrical conditions and no
more. Thus a circle is determined when three points on it
are given, or when it is required to touch three straight
lines.
145. To find the equation to the circle which is described on the
line joining the points {x-^ , y-^) and (x^ , 2/2) ^^ diameter.
Let A be the point (a;^, y-^) and B be the point {x^, 2/2) » ^^^ ^^^ ^^^
coordinates of any point P on the circle be h and k.
The equation to AP is (Art. 62)
y-yi=j^i^-^i) (1).
and the equation to BP is
2/ -2/2=^' (^-^2) (2).
But, since APB is a semicircle, the angle APB is a right angle,
and hence the straight Unes (1) and (2) are at right angles.
122
COOKDINATE GEOMETRY.
Hence, by Art. 69, we have
^-Vi _ h^ljh^ _ 1
i.e. (h-x-^){h-x^) + {k-yj}{Jc-y2) = 0.
But this is the condition that the point {h, k) may lie on the curve
whose equation is
{X - x^) {x - x^) + {y- y^) {y - y^) = 0.
This therefore is the required equation.
146. Intercepts made on the axes by the circle whose equation is
ax^ + ay^ + 2gx + 2fy + c = (1).
The abscissae of the points where the circle (1) meets the axis of x,
i.e. y = 0, are given by the equation
ax^ + 2gx + c = (2).
The roots of this equation being Xj^ and x^ ,
we have
^ + •^2
29
a '■
and
(Art. 2.
Hence
A-^A^ = x^ -x-^= J{x-^-\-x^^ - 4x-^i
V a2
= 2
a a
Again, the roots of the equation (2) are both imaginary if g^<ac.
In this case the circle does not meet the axis of x in real points, i.e.
geometrically it does not meet the axis of x at all.
The circle will touch the axis of x if the intercept A-^A^, be just
zero, i.e. ii g^ = ac.
It will meet the axis of x in two points lying on opposite sides of
the origin if the two roots of the equation (2) are of opposite signs,
i. e. if c be negative.
147. Ex.1. Find the equation to the circle which passes through
the points (1, 0), (0, - 6), and (3, 4).
Let the equation to the circle be
x^ + y^ + 2gx + 2fy + c = (1).
Since the three points, whose coordinates are given, satisfy this
equation, we have
l + 2^ + c = (2),
36-12/+c=0 (3),
and 25 + 6^ + 8/+c = (4).
EXAMPLES. 123
Subtracting (2) from (3) and (3) from (4), we have
2^ + 12/= 35,
and 65f + 20/=ll.
Hence /=*/- and g=z -iJ-.
Equation (2) then gives c = ^-.
Substituting these values in (1) the required equation is
4a;2 + 4i/2 - 142a; + 47a; + 138 = 0.
Ex. 2. Find the equation to the circle which touches the axis ofy
at a distance + 4 from the origin and cuts off an intercept 6 from the
axis of X.
Any circle is x^ + y'^ + 2gx + 2fy + c = 0.
This meets the axis of y in points given by
y^ + 2fy + c = 0.
The roots of this equation must be equal and each equal to 4, so
that it must be equivalent to {y - 4)^ = 0.
Hence 2/= -8, and c = 16.
The equation to the circle is then
x'^ + y^ + 2gx-8y + 16 = 0.
This meets the axis of x in points given by
x'^ + 2gx + l&:=0,
i.e. at points distant
-g+slT^^ and -g-Jg^lJG,
Hence Q = 2^g^-ie.
Therefore ^= ±5, and the required equation is
x^ + y^±10x-8y + U = 0.
There are therefore two circles satisfying the given conditions.
This is geometrically obvious.
EXAMPLES. XVII.
Find the equation to the circle
1. Whose radius is 3 and whose centre is ( - 1, 2).
2. Whose radius is 10 and whose centre is ( - 5, - 6).
3. Whose radius is a + b and whose centre is {a, - h).
4. Whose radius is Ja^ - b'^ and whose centre is { - a, - 6).
Find the^ coordinates of the centres and the radii of the circles
whose equations are
5. x^ + y^-ix-8y = ^l. Q, Sx'^ + Sz/ - 5x- Qy + 4 = 0.
124 COORDINATE GEOMETRY. [ExS.
7. x^ + y^=k{x + k). 8. x^ + y'^ = 2gx-2fy.
9. Jl+m"^ {x^ + if) - 2cx - 2mcy = 0.
Draw tlie circles whose equations are
10. x^ + y^ = 2ay. H, 3x^ + 3y^=4x.
12. 5x'^ + 5y^=2x + 3y.
13. Find the equation to the circle which passes through the
points (1, - 2) and (4, - 3) and which has its centre on the straight
line 305 + 42/ = 7.
14. Find the equation to the circle passing through the points
(0, a) and (6, h), and having its centre on the axis of x.
Find the equations to the circles which pass through the points
15. (0, 0), (a, 0), and (0, 6). 16. (1, 2), (3, -4), and (5, -6).
17. (1,1), (2,-1), and (3, 2). 18. (5, 7), (8, 1), and (1, 3).
19. («, &)> («j -&)> and(a + 6, a-b).
20. ABGB is a square whose side is <x; taking AB and AD as
axes, prove that the equation to the circle circumscribing the square is
x^-\-y^ = a{x-\-y).
21. Find the equation to the circle which passes through the
origin and cuts off intercepts equal to 3 and 4 from the axes.
22. Find the equation to the circle passing through the origin
and the points (a, 6) and (6, a). Find the lengths of the chords that
it cuts off from the axes.
23. Find the equation to the circle which goes through the origin
and cuts off intercepts equal to h and h from the positive parts of the
axes.
24. Find the equation to the circle, of radius a, which passes
through the two points on the axis of x which are at a distance & from
the origin.
Find the equation to the circle which
25. touches each axis at a distance 5 from the origin.
26. touches each axis and is of radius a,
27. touches both axes and passes through the point ( - 2, - 3).
28. touches the axis of x and passes through the two points
(1, -2) and (3, -4).
29. touches the axis of y at the origin and passes through the
point (6, c).
XVII.] TANGENT TO A CIRCLE. 125
30. touches the axis of a; at a distance 3 from the origin and
intercepts a distance 6 on the axis of y.
31. Points (1, 0) and (2, 0) are taken on the axis of x, the axes
being rectangular. On the line joining these points an equilateral
triangle is described, its vertex being in the positive quadrant. Find
the equations to the circles described on its sides as diameters.
32. If 1/ = mx be the equation of a chord of a circle whose radius is
a, the origin of coordinates being one extremity of the chord and the
axis of X being a diameter of the circle, prove that the equation of a
circle of which this chord is the diameter is
(1 + m2) (£c2 + 1/2) -2a{x + my) = 0.
33. Find the equation to the circle passing through the points
(12, 43), (18, 39), and (42, 3) and prove that it also passes through
the points ( - 54, - 69) and ( - 81, - 38).
34. Find the equation to the circle circumscribing the quadrilateral
formed by the straight lines
2x + 3y = 2, dx-2y = 4:, x + 2y = 3, and 2x-y = B.
35. Prove that the equation to the circle of which the points
{x^ , y^) and {x^ , y^) ^^^ the ends of a chord of a segment containing an
angle 6 is
{x - x^) {x - x^) + (y - yj) (y - y^)
± cot 9 [{x - Xj) {y - 2/2) - {x -x^) (y - y^)] = 0.
36. Find the equations to the circles in which the line joining the
points {a, h) and {b, - a) is a chord subtending an angle of 45° at any
point on its circumference.
148. Tangent. Euclid in his Book III. defines the
tangent at any point of a circle, and proves that it is always
perpendicular to the radius drawn from the centre to the
point of contact.
From this property may be deduced the equation to the
tangent at any point (x\ y') of the circle x^ ^-y^ — o?.
For let the point P (Fig. Art. 139) be the point
(a;', y').
The equation to any straight line passing through T is,
by Art. 62,
y~ y' = m, {x — x) (1).
Also the equation to OP is
3/ = |* (2).
126 COORDINATE GEOMETRY.
The straight lines (1) and (2) are at right angles, i.e. the
line (1) is a tangent, if
mx^ = -l, (Art. 69)
I. e. II m = > .
y
Substituting this value of m in (1), the equation of the
tangent at {x\ y) is
y-y =--f{^-^h
t.e. xx+yy— x'^ + y (oj.
But, since {x, y) lies on the circle, we have x^ + y"^ — a^,
and the required equation is then
XX' + yy' = a2.
149. In the case of most curves it is impossible to
give a simple construction for the tangent as in the case of
the circle. It is therefore necessary, in general, to give a
different definition.
Tangent. Def. Let F and Q be any two points, near
to one another, on any curve.
Join TQ \ then FQ is called a
secant.
The position of the line PQ when
the point Q is taken indefinitely close
to, and ultimately coincident with, the
point F is called the tangent at F. X ^
The student may better appreciate ^^--,___^^
this definition, if he conceive the curve
to be made up of a succession of very small points (much
smaller than could be made by the finest conceivable drawing
pen) packed close to one another along the curve. The
tangent at F is then the straight line joining F and the
next of these small points.
150. To find the equation of the tangent at the imint
{x\ y) of the circle x^ + y^ = a^.
EQUATION TO THE TANGENT. 127
Let P be the given point and Q a point (x\ y") lying on
the curve and close to P.
The equation to FQ is then
y-2/' = f^(»-»^') (1)-
Since both (cc', y) and {x\ y") lie on the circle, we have
x'^ + y'^ = a\
and x"^ + y'"^ = a^.
By subtraction, we have
x"^-x" + y"'-y" = 0,
i. e. {x" - x') {x" + x') + (y" - y') (y" + y') ^ 0,
// / It , r
y —y X + x
X - X y + y
Substituting this value in (1), the equation to FQ is
y-y'=-^^'(''-'') (^)-
Now let Q be taken very close to F, so that it ulti-
mately coincides with P, i, e. put x" = x and y" = y.
Then (2) becomes
2x
y-y' = -7y-A^-^\
^y
i. e. yy' + xx' — x'^ + y'^ = a^.
The required equation is therefore
xx' + yy' = a2 (3).
It will be noted that the equation to the tangent
found in this article coincides with the equation found
from Euclid's definition in Art. 148.
Our definition of a tangent and Euclid's definition there-
fore give the same straight line in the case of a circle,
151. To obtain the equation of the tangent at any point
(x'y y') lying on the circle
x^ + y'^ + 2gx + 2fy + c = 0.
128 COORDINATE GEOMETRY.
Let P be the given point and Q a point (x'\ y") lying on
the curve close to P.
The equation to PQ is therefore
2/-2/' = C^!(«-x') (1).
Since both (x\ y) and (cc", y") lie on the circle, we have
x'^ + y'^ + 'lgx' + 2fy' + C = (2),
and x"^ + y"' + 2gx" + 2fy" + c^^ (3).
By subtraction, we have
^- _ x'^ + y"^ _ y- + 2g (x" - x') + 2/(y" - y') = 0,
i. e. (x" - x) {x" + x' + 2g) + (y" - y') {y" + y' + 2/) = 0,
y —y X + X •{• Zg
X - X y +y +2/
Substituting this value in (1), the equation to PQ be-
comes
f X + X + Ag . ,. / , V
Now let Q be taken very close to P, so that it ultimately
coincides with P, i. e. put x" — x and y" = y'.
The equation (4) then becomes
^. e. y {y +/) + 0^(03' + ^) = y' {y' +f)+x {x' + g)
to fO / /• /
= a:^ + 2/- + p'aj +fy
= -9^' -fy -(^i
by (2).
This may be written
XX' + yy' + g (x + X') + f (y + y) + c = O
which is the required equation.
152. The equation to the tangent at (x', y') is there-
fore obtained from that of the circle itself by substituting
XX for x^y yy' for y^, x ■{■ x' for 2a;, and y -V y for 2y.
INTERSECTIONS OF A STRAIGHT LINE AND A CIRCLE. 129
This is a particular case of a general rule Avhich will be
found to enable us to write down at sight the equation to
the tangent at ix\ y') to any of the curves with which we
shall deal in this book.
153. Points of intersection, in general, of the straight
line
y^mx -\- G (1),
with the circle x' + y^ = d^ (2).
The coordinates of the points in which the straight line
(1) meets (2) satisfy both equations (1) and (2).
If therefore we solve them as simultaneous equations
we shall obtain the coordinates of the common point or
points.
Substituting for y from (1) in (2), the abscissae of the
required points are given by the equation
x^ + {mx + c)^ = a^,
i.e. x^ (1 + m^) + 2mcx + c^ — «- = (3).
The roots of this equation are, by Art. 1, real, coinci-
dent, or imaginary, according as
(2mc)^ — 4 (1 + m^) (c^ — a^) is positive, zero, or negative,
i.e. according as
a? {\ -\- m^) — c^ is positive, zero, or negative,
i.e. according as
c^ is < = or > a^ (1 + m^).
In the figure the lines marked I, II, and III are all
parallel, i.e. their equations all have the same "r/i."
L. 9
130 COORDINATE GEOMETRY.
The straight line I corresponds to a value of c^ which
is <a^ (1 + oYv^) and it meets the circle in two real points.
The straight line III which corresponds to a value of c^,
> a^ (1 + m^), does not meet the circle at all, or rather, as in
Art. 108, this is better expressed by saying that it meets
the circle in imaginary points.
The straight line II corresponds to a value of c', which
is equal to a^ (1 + m^), and meets the curve in two coincident
points, i.e. is a tangent.
154. We can now obtain the length of the chord inter-
cepted by the circle on the straight line (1). For, if x^ and
X2 be the roots of the equation (3), we have
2mc _, c^ — cC-
^ ^ 1 + m" ^ ^ 1 + m^
Hence
2
^1 — ^2 = Jip^i + ^2)^ — ^x^x^ = -z "2 Jm^c^ - {(f—a^) (1 + m^)
J. "r 7Yh
1 +m^
If 2/1 and 2/2 ^^ t^6 ordinates of Q and R we have, since
these points are on (1),
2/1 — 2/2 = {mx-^ + <^) "~ (^^2 + c) = rri {x^ — x^.
Hence
QR = J(yi - y^Y + K - oc^y = Jl+ m' (x^^ - x.^
y
a^ (1 + m^) — c^
1 + 7n^
In a similar manner we can consider the points of inter-
section of the straight line y = mx + k with the circle
a;^ + 2/2 + 2yx + 2fy + c = 0.
155. The straight line
y = mx + ajl + m^
is always a. tangent to the circle
2 Q o
X + y — a .
EQUATION TO ANY TANGENT. 131
As in Art. 153 the straight line
y = mx + G
meets the circle in two points which are coincident if
But if a straight line meets the circle in two points
which are indefinitely close to one another then, by Art.
149, it is a tangent to the circle.
The straight line y = mx + c is therefore a tangent to the
circle if
G^a J\ + 772^,
i.e. the equation to any tangent to the circle is
y = mx + a Vl + m2 (1).
Since the radical on the right hand may have the + or —
sign prefixed we see that corresponding to any value of in
there are two tangents. They are marked II and IV in
the figure of Art. 153.
156. The above result may also be deduced from the equation
(3) of Art. 150, which may be written
x' a2
y= --«+- (1).
y y
x'
Put — , = 'm, so that x'= -my' , and the relation .r'- + ?/'2 = a^ gives
y'^{m^+l) = a^, i.e. —= /Jl+m^.
The equation (1) then becomes
y = mx + afjl + m\
This is therefore the tangent at the point whose coordinates are
~ma . a
and
Vi + w2 7i +
W''
157. If we assume that a tangent to a circle is always perpen-
dicular to the radius vector to the point of contact, the result of
Art. 155 may be obtained in another manner.
For a tangent is a line whose perpendicular distance from the
centre is equal to the radius.
9—2
132 COORDINATE GEOMETRY.
The straight line y=mx + c will therefore touch the circle if the
perpendicular on it from the origin be equal to a, i.e. if
i.e. if c = a /sjl + m^.
This method is not however applicable to any other curve besides the
circle.
158. Ex. Find the equations to the tangents to the circle
x'^ + y^-6x + Ay = 12
which are parallel to the straight line
4a;+3i/ + 5 = 0.
Any straight line parallel to the given one is '
4a: + 32/ + (7 = (1).
The equation to the circle is
(a; -3)2 + (2/ + 2)2 =52.
The straight line (1), if it be a tangent, must be therefore such
that its distance from the point (3, - 2) is equal to ±5.
Hence ^l7^±f=:±5 (Art. 75),
V4H32
so that (7= -6=t25 = 19or -31.
The required tangents are therefore
4a; + 32/ + 19 = and 4;r + 3?/-31 = 0.
159. Normal. Def. The normal at any point P of
a curve is the straight line which passes through P and is
perpendicular to the tangent at P.
To find the equation to the normal at the j^oint (x', y'^ of
(1) the circle
X- + 2/ = a%
and (2) the circle
X- + y^ + Igx + 2/2/ + c = 0.
(1) The tangent at (.t', 3/') is
XX + yy — a^,
X a^
i.e. y^ ,x + — .
y y
THE NORMAL TO THE CIRCLE. 133
The equation to the straight line passing through {x\ y)
perpendicular to this tangent is
y-y' ^m{x- x),
where m x (- %] = - 1, (Art. 69),
y'
X
The required equation is therefore
y-y ^-'(^-^-^)' .
i. e. X y — xy = 0.
This straight line passes through the centre of the circle
which is the point (0, 0).
If we assume Euclid's propositions the equation is at once
written down, since the normal is the straight line joining
(0, 0) to («;', 2/').
(2) The equation to the tangent at {x ^ y) to the circle
'£ + i/-^ + ^yx + %fy + c —
X + CI gx + fu + c , ^ T ^ ^ V
IS y = -- — -.x-^- /-^^ . (Art. 151.)
The equation to the straight line, passing through the
point (a;', ?/') and perpendicular to this tangent, is
y-y =m{x- x'),
where m x f-%^\ - - 1, (Art. 69),
y+f
I.e. m^—, — - .
x + g
The equation to the normal is therefore
y-y'-^K~^{^-x),
X + g ^ '
i. e. y {x' +g)-x{y +/)+ fx - gy' = 0.
134 COORDINATE GEOMETRY.
EXAMPLES. XVIII.
Write down the equation of the tangent to the circle
1. a;2 + 7/2 -3a; + 10?/ = 15 at the point (4, -11).
2. 4^2 + iy^ - 16a; + 24y -^ 117 at the point ( - 4, - -V) .
Find the equations to the tangents to the circle
3. a;2 + 2/^ = 4 which are parallel to the line a; + 2^ + 3 = 0.
4. -JC' + 2/^ + 2</a; + 2fy + c = which are parallel to the line
a; + 22/-6 = 0.
5. Prove that the straight line y = x + cJ2 touches the circle
x'^-\-y'^ = c-\ and find its point of contact.
6. Find the condition that the straight line ex ~hy + 1x^ = may
touch the circle x'^ + y^ = ax + by and find the point of contact.
7. Find whether the straight line x + y = 2 + J 2, touches the circle
x- + y^-2x-2y + l = Q.
8. Find the condition that the straight line ^x + ^y = k may
touch the circle x'^-\-y'^ — lQx,
9. Find the value oip so that the straight line
cccos a+?/sina-p =
may touch the circle
a;2 + 7/2 - 2aa; cos a - 2hy sin a - a^ sin^ a — 0.
10. Find the condition that the straight line Ax-\-By + G — Q may
touch the circle
{x-af+{y-hf=c^.
11. Find the equation to the tangent to the circle x--\-y'^ — a?
which
(i) is parallel to the straight line y = mx + c,
(ii) is perpendicular to the straight line y = mx + c,
(iii) passes through the point (6, 0),
and (iv) makes with the axes a triangle whose area is a-.
12. Find the length of the chord joining the points in which the
straight line
X y ,
a
meets the circle x'^ + y^ = r".
13. Find the equation to the circles which pass through the origin
and cut off equal chords a from the straight lines y — x and y—-x.
[EXS. XVIII.] EXAMPLES. 135
14. Find the equation to the straight lines joining the origin to
the points in which the straight Hne y = rnx + c cuts the circle
x^ + y'^ = 2ax + 2by .
Hence find the condition that these points may subtend a right
angle at the origin.
Find also the condition that the straight line may touch the
circle.
Find the equation to the circle which
15. has its centre at the point (3, 4) and touches the straight line
5x + 12y = l.
16. touches the axes of coordinates and also the line
a
the centre being in the positive quadrant.
17. has its centre at the point (1, - 3) and touches the straight
Hne 2a: -i/- 4 = 0.
18. Find the general equation of a circle referred to two perpen-
dicular tangents as axes.
19. Find the equation to a circle of radius /• which touches the
axis of 2/ at a point distant h from the origin, the centre of the circle
being in the positive quadrant.
Prove also that the equation to the other tangent which passes
through the origin is
{r^-h'^)x + 2rhy = 0.
20. Find the equation to the circle whose centre is at the point
(a, p) and which passes through the origin, and prove that the
equation of the tangent at the origin is
21. Two circles are drawn through the points {a, 5a) and (4a, a)
to touch the axis of y. Prove that they intersect at an angle tan^^ y* .
22. A circle passes through the points ( - 1, 1), (0, 6), and (5, 5).
Find the points on this circle the tangents at which are parallel to the
straight line joining the origin to its centre.
160. To shew that from any i^oint there caii he drawn
two tangents^ real or imaginary^ to a circle.
Let the equation to the circle be '3t? ■\- y"^ — «', and let the
given point be (ajj, y^. [Fig. Art. 161.]
The equation to any tangent is, by Art. 155,
y — mx + a J\ + rn?.
136 COORDINATE GEOMETllY.
If this pass through the given point (xj^, y^ we have
v/i = mx'i + aj\ + m^ (1).
This is the equation which gives the values of m corre-
sponding to the tangents which pass through (x-^^^ y^).
Now (1) gives
?/i — jiid\ — a Jl + m^,
i.e. y^ — 2rax-^y-^ + m^x^ = «- + ahn^^
i. e. m^ {xj^ — a^) — 2mx^y^ + y-^^ — a^ = (2).
The equation (2) is a quadratic equation and gives
therefore two values of ?m (real, coincident, or imaginary)
corresponding to any given values of x^^ and y^. For each
of these values of 7?i we have a corresponding tangent.
The roots of (2) are, by Art. 1, real, coincident or
imaginary according as
(2xi?/i)^ — 4 (.r/ — a"^) (2/1^ — a^) is positive, zero, or negative,
i. e. according as
a^ (— a"^ + Xj^ + y^) is positive, zero, or negative,
i. e. according as x-^ + y^ =■- a?.
If x^ + y^> a^, the distance of the point (x^^, y^) from
the centre is greater than the radius and hence it lies outside
the circle.
If x{- + yi = a?, the point {x^ , y^ lies on the circle and
the two coincident tangents become the tangent at (x-^^ , y^.
If x-^ + y^ <«^, the point {x-^^ y^ lies within the circle,
and no tangents can then be geometrically drawn to the
circle. It is however better to say that the tangents are
imaginary.
161. Chord of Contact. Def. If from any point
T without a circle two tangents TP and TQ be drawn to
the circle, the straight line PQ joining the points of
contact is called the chord of contact of tangents from T.
To find the equation of the chord of contact of tangents
drawn to the circle x^ + y"^ — a^ from the external point
(^1, 2/i)-
POLE AND POLAR.
137
•(3),
.(4).
Let T be the point (x^, y^), and F and ^ the points
(oj'j 2/') and (a?", y") respectively.
The tangent at F is
XX +1/1/' — a-^ (1),
and that at Q is
xx" + yy" —d^ (2).
Since these tangents pass through
T^ its coordinates {x-^ , y^ must satisfy
both (1) and (2).
Hence x-^ + y^y — a?
and x^od' + y-^y" = cir
The equation to FQ is then
xxi + yyi=a2 (5).
For, since (3) is true, it follows that the point {x ^ y),
i.e. F, lies on (5).
Also, since (4) is true, it follows that the point (x\ y")^
i.e. Q, lies on (5).
Hence both F and Q lie on the straight line (5), i.e.
(5) is the equation to the required chord of contact.
If the point {x^, y^) lie within the circle the argument
of the preceding article will shew that the line joining the
(imaginary) points of contact of the two (imaginary)
tangents drawn from (x^ , y^) is xx^ + yy^ — a^.
We thus see, since this line is always real, that we may
have a real straight line joining the imaginary points of
contact of two imaginary tangents.
162. Pole and Polar. Def. If through a point
F (within or without a circle) there be drawn any straight
line to meet the circle in Q and F, the locus of the point of
intersection of the tangents at Q and F is called the polar
of F ; also F is called the pole of the polar.
In the next article the locus will be proved to be a
straight line.
138
COORDINATE GEOMETRY.
163. To Jincl the equation to the 'polar of the point
(^ij 2/1) '^^ih respect to the circle x^ ■\-y^ — d^.
Let QR be any chord drawn through P and let the
tangents at Q and R meet in the point :Z^ whose coordinates
are (A, k).
Hence QR is the chord of contact of tangents drawn
from the point (h, k) and therefore, by Art. 161, its
equation is xh ■\- yk — o?.
Since this line passes through the point (x-^, y^ we
have
x-Ji + y^ — a" (1).
Since the relation (1) is true it follows that the
variable point (A, k) always lies on the straight line whose
equation is
xxi + yyi = a2 (2).
Hence (2) is the polar of the point (x^, y^).
In a similar manner it may be proved that the polar of
(x'l, i/i) with respect to the circle
o(^ + y^ + 2gx + 2fy + c =
is xx^ -^yvi + g (x + xi) +/ (y + yi) + c=^ 0.
164. The equation (2) of the preceding article is the
same as equation (5) of Art. 161. If, therefore, the point
{xi, 2/i) be without the circle, as in the right-hand figure,
the polar is the same as the chord of contact of the real
tangents drawn through (x^^, y^).
If the point (x^, y^) be on the circle, the polar coincides
with the tangent at it. (Art. 150.)
GEOMETRICAL CONSTRUCTION FOR THE POLAR. 139
If the point (x^, y-f) be within the circle, then, as in
Art. 161, the equation (2) is the line joining the (imaginary)
points of contact of the two (imaginary) tangents that can
be drawn from (x-^ , y^.
We see therefore that the polar might have been
defined as follows :
The polar of a given point is the straight line which
passes through the (real or imaginary) points of contact of
tangents drawn from the given point ; also the pole of any
straight line is the point of intersection of tangents at the
points (real or imaginary) in which this straight line meets
the circle.
165. Geometrical construction for the imlar of a point.
The equation to OP^ which is the line joining (0, 0) to
^.e.
■(!)•
Also the polar of P is
xx^ + yVx — cC"
(2).
By Art. 69, the lines (1) and (2) are perpendicular to
one another. Hence OP is perpendicular to the polar
of P.
Also the length OP - ^x^+y},
140 COORDINATE GEOMETRY,
and the perpendicular, OiV, from upon (2)
Hence the product ON . OP — a^.
The polar of any point P is therefore constructed thus :
Join OP and on it (produced if necessary) take a point N
such that the rectangle ON . OP is equal to the square of
the radius of the circle.
Through N draw the straight line LL' perpendicular to
OP ; this is the polar required.
[It will be noted that the middle point N of any chord LL' lies on
the line joining the centre to the pole of the chord.]
166. To Jind the pole of a given line with respect to
any circle.
Let the equation to the given line be
^a; + % + (7-0 (1).
(1) Let the equation to the circle be
and let the required pole be {x-^, y-^.
Then (1) must be the equation to the polar of (oji, y^,
i.e. it is the same as the equation
xx-^ + yVi — cv^-O (2).
Comparing equations (1) and (2), Ave have
J B C '
so that Xt — — yz a^ and y, = — '^ a?.
C ^ C
The required pole is therefore the point
A , B ^
(2) Let the equation to the circle be
a;- + 2/2 + l.yx + 2fy + g=Q.
POLE AND POLAR. 141
If (a?!, 2/1) ^^ *^® required pole, then (1) must be
equivalent to the equation
xx^ +yyi + g{os + x,) +f{y + yi) + c=0, (Art. 1 63),
i.e. x(x^+g)+y{y^+f) + gxj^ +/2/i + c = (3).
Comparing (1) with (3), we therefore have
- sc^ + 9 _ Vi+f^ 9^1 +/2/1 + g
A " B C
By solving these equations we have the values of x^
and 2/1.
Ex. Find the pole of the straight line
9x+2j-28 = (1)
2vith respect to the circle
2x^+2y'^-Sx + 5ij-7 = (2).
If (.Tj, yj) be the required point the line (1) must coincide with the
polar of (x^, 2/i)' whose equation is
2xx^ + 2yy-^-i{x + x^)+^{y+y{j-l = 0,
i.e. x{4x^-d) + y{4:y^ + 5)-3xi + 5y^-U = (3).
Since (1) and (3) are the same, we have
4.'Ci - 3 _ 4|/i + 5 _ -Sxi + 5y^ - 14
9 r~ ~ ^28 ■
Hence Xi = 9yi + 12,
and 3.Ti-117i/i = 126.
Solving these equations we have Xj^ = S and t/^ = - 1, so that the
required point is (3, - 1).
167. If the 2')olar of a point P pass through a point T,
then the polar of T passes through P.
Let P and T be the points {x-^, y^ and (.Tg, 2/2) ^^-
spectively. (Fig. Art. 163.)
The polar of (x^, y^ with respect to the circle
a;^ + 2/2 _ ^2 jg
xx-^ + 2/2/1 = Cb^'
This straight line passes through the point T if
jcoa?! + 2/22/1 = ct^ ••• (!)•
142 COORDINATE GEOMETRY.
Since the relation (1) is true it follows that the point
(x^, 2/1), i.e. P, lies on the straight line xx^-\-yy^ = (]?^ which
is the polar of {x^^ y^)^ i.e. T, with respect to the circle.
Hence the proposition.
Cor. The intersection, T, of the polars of two points,
P and Q, is the pole of the line PQ.
168. To find the length of the tangent that can he
drawn from the point (x^ , y-^ to the circles
(1) a?-^y'^ = a\
and (2) x? + y^ + 2gx + 2/^/ + c = 0.
If T be an external point (Fig. Art. 163), TQ a tangent
and the centre of the circle, then TQO is a right angle
and hence
(1) If the equation to the circle be x" ^-y"- — a^, is the
origin, OT'^ = x^- + y^.^ and OQ^ = a^.
Hence TQ" = x^^ + y^- - a\
(2) Let the equation to the circle be
x^ + y'^+ 2gx + 2fy + c = 0,
i.e. (x + gf-v{y+fY = f+f'-c.
In this case is the point (— g, —f) and
OQ^ — (radius)^ = g'^ +f^ — c.
Hence OT^ = [x, - (- g)Y + [y, - (-/)]' (Art. 20).
Therefore TQ^ ^ {x, + gf + (y, +ff - (f +p - c)
= <^ + Vi+'^g^i + 2/J/i + c.
In each case we see that (the equation to the circle
being written so that the coefficients of x^ and y" are each
unity) the square of the length of the tangent drawn to the
circle from the point {x^, y^) is obtained by substituting x^
and 2/1 for the current coordinates in the left-hand member
of the equation to the circle.
*169. To find the equation to the pair of tangents that
can he drawn from the point (x-^^ y^) to the circle x- + y^ — d^.
PAIR OF TANGENTS FROM ANY POINT. 143
Let (A, k) be any point on either of the tangents from
(^1, Vi)'
Since any straight line touches a circle if the perpen-
dicular on it from the centre is equal to the radius, the
perpendicular from the origin upon the line joining (x^, y^)
to (A, k) must be equal to a.
The equation to the straight line joining these two
points is
^ — Vi /
i.e. yQ^~~ ^'i) ~ ^ (^ ~ 2/1) + ^^1 ~~ ^^Ui — ^■
__ kx^ — hy^
Hence , ^ = a,
J(h-x,f+(k-y,y
so that (kx^ — hy^y = a^ [{h — oo^y + {k — 2/1)^]-
Therefore the point (h, k) always lies on the locus
{x^y - xy^f = a" [{x - x^y + {y- y,y] (1).
This therefore is the required equation.
The equation (1) may be written in the form
= 2xyx^yi — 2a^xXj^ — 2a^yyi,
i.e. (x^ + y'^- CL^) {^i + y\ — «^) = x'^X{' + y^y^ + a^ + 2xyx{y^
— 2a^xXi — 2c^yy-^ — {xx^ + yy^ — a^y (2).
#170. In a later chapter we shall obtain the equation to the pair
of tangents to any curve of the second degree in a form analogous
to that of equation (2) of the previous article.
Similarly the equation to the pair of tangents that can be
drawn from {x-^, y-^ to the circle
If the equation to the circle be given in the form
a;2 + 2/2 + 2,9ra; + 2/?/ + c =
the equation to the tangents is, similarly,
{x' + ^2 + ^g^ + 2/y + c) {x^ + y^ + 2^a;i + 2jy^ + c)
= \xx^^yy^^g{x\x^^f{y^ry^) + cf (2).
144 COORDINATE GEOMETRY.
EXAMPLES. XIX.
Find the polar of the point
1. (1, 2) with respect to the circle a;- + y^ = 7.
2. (4, - 1) with respect to the circle 2x" + 2y^=ll,
3. (-2,3) with respect to the circle
x'^ + y^-4x-&y + 5 = 0.
4. (5, - I) with respect to the circle
Sx^ + Sy^-1x + 8y-9 = 0.
5. (a, - 6) with respect to the circle
x^ + y^ + 2ax - 2by + a^ - b- = 0.
Find the pole of the straight line
6. x + 2y = l with respect to the circle x^ + y^=5.
7. 2x-y = G with respect to the circle 5x^ + oy^ =9.
8. 2x + y + 12 = with respect to the circle
x^ + y^-4x + By-l=0.
9. 48;r - 54?/ + 53 = with respect to the circle
Sx^ + 3y^ + 5x-7y + 2 = 0.
10. ax + by + 3a^ + 36^ = with respect to the circle
x^ + y^ + 2ax + 2by = a^ + b^.
11. Tangents are drawn to the circle x^ + y^ = 12 at the points
where it is met by the circle x^ + y^ - 5x + Sy ~ 2 = ; find the point of
intersection of these tangents.
12. Find the equation to that chord of the circle x'^ + y^ = 81 which
is bisected at the point ( - 2, 3), and its pole with respect to the circle.
13. Prove that the polars of the point (1, - 2) with respect to the
circles whose equations are
x^ + y^ + Qy + 5 = and x'^ + y^ + 2x + 8y + 5 =
coincide ; prove also that there is another point the polars of which
with respect to these circles are the same and find its coordinates.
14. Find the condition that the chord of contact of tangents from
the point (x', y') to the circle x^+y^=a^ should subtend a right angle
at the centre.
15. Prove that the distances of two points, P and Q, each from
the polar of the other with respect to a circle, are to one another
inversely as the distances of the points from the centre of the circle.
16. Prove that the polar of a given point with respect to any one
of the circles x^-\-y'^-2kx-\-c'^ = 0, where k is variable, always passes
through a fixed point, whatever be the value of A\
[EXS. XIX.] POLAR EQUATION TO THE CIRCLE.
145
17. Tangents are drawn from the point {h, k) to the circle
x^ + y^ = aP; prove that the area of the triangle formed by them
and the straight line joining their points of contact is
ajJi^ + Ji^-aY
h^ + k^
Find the lengths of the tangents drawn
18. to the circle '2x^ + 2y'^—S from the point ( - 2, 3).
19. to the circle 3x^ + 3y^ -7x-6y = 12 from the point (6, - 7).
20. to the circle x^ + f' + Ihx - 36^ = from the point
(a + &, a-h).
21. Given the three circles
3ic2 + 37/-36a; + 81 = 0,
and a:2^.,y2_i6^_12y + 84 = 0,
find (1) the point from which the tangents to them are equal in
length, and (2) this length.
22. The distances from the origin of the centres of three circles
ic2-hy"^-2Xic = c^ (where c is a constant and X a variable) are in
geometrical progression ; prove that the lengths of the tangents drawn
to them from any point on the circle a;"^ + ^- = c-are also in geometrical
progression.
23. Find the equation to the pair of tangents drawn
(1) from the point (11, 3) to the. circle xr'-\-y^ — ^^,
(2) from the point (4, 5) to the circle
2x-- + 2i/3 - 8a; + 12?/ + 21 r= 0.
171. To jincl the general equation of a circle referred
to polar coordinates.
Let be the origin, or pole, OX the initial line, G the
centre and a the radius of the
circle.
Let the polar coordinates of C
be R and a, so that 00 — M and
L XOC = a.
Let a radius vector through
at an angle 6 with the initial line
cut the circle in P and Q. Let
OP, or OQ, be r.
L.
10
146 COOKDINATE GEOMETRY.
Then {Trig. Art. 164) we have
CP- ^ OC'^ -r OP' -20C. OP cos C OF,
i.e. a''^P^ + r'-2Prcos(e~a),
i.e. r'- -2Rr cos {0 - a) + R"" - a" ^0 (1).
This is the required polar equation.
172. Particular cases of the general equation inpolar coordinates.
(1) Let the initial line be taken to go through the centre G. Then
a = 0, and the equation becomes
r2 - 2Rr cos ^ + E^ - a^ = 0.
(2) Let the pole be taken on the circle, so that
B=OG = a.
The general equation then becomes
^^M r^-2arcos{d-a) = 0,
i.e. r=2acos{d - a).
(3) Let the pole be on the circle and also let the initial line pass
through the centre of the circle. In this case
a=0, and R — a.
The general equation reduces then to the
simple form r = 2acos^.
This is at once evident from the figure. 0|
For, if OCA be a diameter, we have
0P=0^ cos ^,
i.e. r=2aQO8 0.
173. The equation (1) of Art. 171 is a quadratic
equation which, for any given value of 6, gives two
values of r. These two values in the figure are OP and
OQ.
If these two values be called r-^ and rg, we have, from
equation (1),
7\r^ = product of the roots — E^ — a"^,
i.e. OP.OQ^P'-a\
The value of the rectangle OP . OQ is therefore the
same for all values of 0. It follows that if we drew any
other line through to cut the circle in P^ and Q^ we
should have OP . OQ = OP^ .OQ^.
This is Euc. iii. 36, Cor.
POLAR EQUATION TO THE TANGENT. 147
174. Find the equation to the chord joining the points on the circle
r — 2a cos 6 whose vectorial angles are d^ and 6^, and deduce the equation
to the tangent at the point 6^.
The equation to any straight line in polar coordinates is (Art. 88)
^ = rcos((9-a) (1).
If this pass through the points (2a; cos ^j^, 6^ and (2asin^2» ^2)' ^^^
have
2a cos d-^ cos (^j-a)=^ = 2acos ^.3 cos {9.^~a) (2).
Hence cos (2^^ - a) + cos a = cos (2^., - a) + cos a,
i.e. 2^1 -a= -(2^0 -a),
since d^ and ^2 ^^^ iiot» i'^ general, equal.
Hence a — d-^ + d.^^
and then, from (2), j? = 2a cos 6-^ cos d.-^.
On substitution in (1), the equation to the required chord is
r cos (^ - ^1 - ^2) = 2a cos ^1 cos ^2 (3).
The equation to the tangent at the point d^ is found, as in
Art. 150, by putting 6^ = 6^ in equation (3).
We thus obtain as the equation to the tangent
rcos (^-2^^) — 2acos2^j^. •
As in the foregoing article it could be shewn that the equation to
the chord joining the points d^ and 6^ on the circle r= 2a cos {d - 7) is
r cos {d -9-^- 6., + 7] — 2a cos {9-^ - 7) cos (9.^ - 7)
and hence that the equation to the tangent at the point ^^ is
r cos (^ - 2^j + 7) = 2a cos2 (^j - 7).
EXAMPLES. XX.
1. Find the coordinates of the centre of the circle
r=A cos 9 + B sin 9.
2. Find the polar equation of a circle, the initial line being a
tangent. What does it become if the origin be on the circumference?
3. Draw the loci
(1) r=a; (2) r = a sin ^; (3) r = ttcos^; (4) r = asec^;
(5) r = acos(^-a); (6) r = asec(^-a).
4. Prove that the equations r = acos(^-a) and r—b sin (9 -a)
represent two circles which cut at right angles.
5. Prove that the equation r^ cos ^- a;* cos 2^ -2a- cos ^ =
represents a straight line and a circle.
10—2
148 COORDINATE GEOMETRY. [Exs. XX.]
6. Find the polar equation to the circle described on the straight
line joining the points {a, a) and (&, /3) as diameter.
7. Prove that the equation to the circle described on the straight
line joining the points (1, 60°) and (2, 30°) as diameter is
r^-r [cos {d - 60°) + 2 cos {0 - 30°)] + ^3 = 0.
8. Find the condition that the straight line
- = acos d + h sin^
r
may touch the circle r = 2c cos d.
175. To find the general equation to a circle referred to
oblique axes which meet at an angle to.
Let C be the centre and a the radius of the circle. Let
the coordinates of C be (h, k) so
that if CM, drawn parallel to the
axis of 2/, meets OX in M, then
OM=h and MC=.k.
Let P be any point on the
circle whose coordinates are x and
y. Draw PN, the ordinate of P,
and CL parallel to OX to meet
PNinL.
Then CL = MN = OiV - OM =- x - h,
and LP^NP-NL^NP -MG^y- k.
Also z CLP - z ONP= 180° - z PNX = 180° - tu.
Hence, since CL^- + LP^ - WL . LP cos CLP = a\
we have (x - h)2 + (y - k)2 + 2 (x -- h) (y - k) cosco = 3?,
i.e. x^ + y^ + 2xy cos w - 2x [h + k cos w) ~2y ik + h cos w)
+ JiT + k^ + 2M COS w — or.
The required equation is therefore found.
176. As in Art. 142 it may be shewn that the
equation
aj- + ^xy cos oi-\- y" + '2gx + 2fy + c -
represents a circle and its radius and centre found.
OBLIQUE COORDINATES. J 49
Ex. If the axes he inclined at 60°, prove that the equation
x^ + xy+i/-^x-5y -2 — (1)
represents a circle and find its centre and radius.
If w be equal to 60°, so that cosw=|, the equation of Art. 175
D6COII1GS
x'^ + xy->ry--x{2h + Tc)-y{2k + h) + h'^ + l'^ + hh = a:^.
This equation agrees with (1) if
2/i + /c = 4 (2),
'ih + h^b (3),
and 7i2 + A;2 + 7i;e-a2= -2 (4).
Solving (2) and (3), we have h = l and k — 2. Equation (4) then
gives
- a2^7t2+F + ;iA; + 2z^9,
so that a = 3.
The equation (1) therefore represents a circle whose centre is the
point (1, 2) and whose radius is 3, the axes being inclined at 60°.
EXAMPLES. XXI.
Find the inclinations of the axes so that the following equations
may represent circles, and in each ease find the radius and centre ;
1. x'-xy-\-y'^~2gx-'ify = 0.
2. x'^ + fjZxy+if-^x-%y + 5 = Q.
3. The axes being inclined at an angle w, find the centre and
radius of the circle
a;2 + 2xy cos w + t/^ - Igx - 2fy = 0.
4. The axes being inclined at 45°, find the equation to the circle
whose centre is the point (2, 3) and whose radius is 4.
5. The axes being inclined at 60°, find the equation to the circle
whose centre is the point ( - 3, - 5) and whose radius is 6.
6. Prove that the equation to a circle whose radius is a and
which touches the axes of coordinates, which are inclined at an angle
to, is
x^ + 2xycoso} + y'^-2a (.T + ^)cot- +a^cot^-=0.
7. Prove that the straight line y — mx will touch the circle
x^ + 2xy cos (o + y^ + 2gx + 2fy + c =
if {g +fm)^ = c (1 + 2m cos o) + m^) .
8. The axes being inclined at an angle w, find the equation to the
circle whose diameter is the straight line joining the points
{x\ y') and {x", y").
150 COORDINATE GEOMETRY.
Coordinates of a point on a circle expressed in
terms of one single variable.
177. If, in the figure of Art. 139, we put the angle
MOP equal to a, the coordinates of the point P are easily
seen to be a cos a and a sin a.
These equations clearly satisfy equation (1) of that
article.
The position of the point P is therefore known when
the value of a is given, and it may be, for brevity, called
" the point a."
With the ordinary Cartesian coordinates we have to
give the values of two separate quantities x and y' (which
are however connected by the relation x' = Ja^ — y"^) to
express the position of a point P on the circle. The
above substitution therefore often simplifies solutions of
problems.
178. To find tlie equation to the straight line joining
two joints, a and y8, 07i the circle xr + y^ = a-.
Let the points be P and Q, and let OA^ be the perpen-
dicular from the origin on the straight line PQ ; then OJV
bisects the angle POQ, and hence
z X0]^= 1 ( z XOP + L XOQ) = H« + ^)-
Also OK^ OP cos NOP = a cos °^^ .
The equation to PQ is therefore (Art. 53),
a + j8 . a+ i8 a — /?
X cos — p-^" + y sm — ^ - — a cos — - .
2 -^ 2 2
If we put P = a we have, as the equation to the tangent
at the point a,
X cos a + y sin a = a.
This may also be deduced from the equation of Art. 150
by putting x' = a cos a and y =cb sin a.
179. If the equation to the circle be in the more
general form
(x - hy + (y-ky^ a', (Art. 1 40),
THE CIRCLE. ONE VARIABLE. 151
we may express the coordinates of F in the form
(A + (X cos a, ^ + ct sin a).
For these values satisfy the above equation.
Here a is the angle LOP [Fig. Art. 140].
The equation to the straight line joining the points a and
^ can be easily shewn to be
/ IN « + yS / 7\ • a + )8 a. — ^
{x — h) cos — ^r^ -^ {y — Ic) sm — ^ — = a cos — ^ ,
and so the tangent at the point a is
{x — K) cos a + (2/ — li) sin a = a.
#180. Ex. Find the four common tangents to the two circles
ox'^ + 5i/ - 22x + iy + 20 = 0,
and 5a;2 + 5?/2 + 22a; -4y- 20 = 0.
The equations may be written
and (x + \^-f+{y-ir^ = SK
Any point on the first circle is (\^- + cos d, -| + sin d).
Any point on the second is ( - V- + 3 cos 0, f + 3 sin 0).
The equations to the tangents at these points are by the last
article
(a;- i^)cos^ + {y + |)sin^ = l (1),
and (a; + -V-)eos0 + (?/-|)sin0 = 3 (2).
These tangents coincide, in which case we have the common
tangents, if
cos ^ _ sin ^ _ - 11 cos ^ + 2 sin ^ - 5
cos <p ~ sin 9f> ~~ 11 cos 0-2 sin 0-15 ^ ^'
From the first pair of these equations we have (f) — dor(p = d + 180°.
If 0=:^, the second pair gives
-llcos^ + 2sin^-5
~ llcos^-2sin^-l"5 '
i.e. llcos^-2sin^ = 5 (4),
i.e. Il-2tan^ = 5^1 + tan2^.
On solving, we have tan ^ = | or - -^^.
152 COOEDINATE GEOMETRY.
Corresponding to tan^ = | we have the values sin ^ = 4, cos^ = f
which satisfy (4) ; we have also the values sin ^= - 1 and cos ^= - f ,
which do not satisfy it.
Corresponding to tan^= --y-> t^ie values which satisfy (4) are
sin^= -ff and cos^=/6.
If = 180°+ ^, the second pair of equations (3) give
- 11 cos ^ + 2 sin ^ - 5
~ -llcos^ + 2sin^-15'
i,e. 11 cos (9 -2 sin ^= -10 (5).
This gives (ll-2tan ^)2= 100(1 + tan2 6>),
so that tan^=-for^.
The corresponding values which satisfy (5) are found to be
sin 0=1, cos0=-| and sin^=-^t., cos0=-||.
The solutions of (3) are therefore
sin0=f, -i^, f, and -^^;
cos = 1, 2^5, -I, and -ff.
On substituting these values in equation (1), the common tangents
are found to be
3a; + 42/ = 10, 7.^-24?/ = 50,
4a;-3i/= 5, and 24.^+ 7?/ = 25.
181. We shall conclude this chapter with some mis-
cellaneous examples on loci.
Ex. I. Find the locus of a point P lohich moves so that its distance
from a given point is always in a given ratio [n : 1) to its distance
from another given point A.
Take as origin and the direction of OA as the axis of x. Let
the distance OA be rt, so that A is the point {a, 0).
If {Xy y) be the coordinates of any position of P we have
OP2=n2.^p2,
i.e. x'^-\-y'^ = n^[{x-a)^-\-y-'\,
i.e. (a;2 + 2/2)(n2-l)-2an2a; + 7i2a2=0 (1).
Hence, by Art. 143, the locus of P is a circle.
Let this circle meet the axis of x in the points C and D. Then OC
and OB are the roots of the equation obtained by putting y equal to
zero in (1).
-.-.- ^ ^ na , ^ ^ na
Hence 00= and 0D= 5.
n + 1 n-1
THE CIRCLE. EXAMPLES. 153
We therefore have
CA= and AD =
n+1 n-1
OG OB
^^^^^ GA=AD = ''-
The points C and D therefore divide the line OA in the given ratio,
and the required circle is on CD as diameter.
Ex. 2. From any 'point on one given circle tangents are drawn to
another given circle ; prove that the locus of the middle point of the
chord of contact is a third circle.
Take the centre of the first circle as origin and let the axis of x
pass through the centre of the second circle. Their equations are
then
a;2+2/2=a2 (1),
and (.'c-c)2 + r/2 = &2 (2),
where a and b are the radii, and c the distance between the centres, of
the circles.
Any point on (1) is {a cos 6, a sin 6) where is variable. Its chord
of contact with respect to (2) is
{x - c) {a cos 9 - c) -{-ya ^m 6 = 1^ (3) .
The middle point of this chord of contact is the point where it is
met by the perpendicular from the centre, viz. the point (c, 0).
The equation to this perpendicular is (Art. 70)
-{x-c)a sin 6 + [a cos 6 -c)y = (4).
Any equation deduced from (3) and (4) is satisfied by the coordi-
nates of the point under consideration. If we eliminate 6 from them,
we shall have an equation always satisfied by the coordinates of the
point, whatever be the value of 6. The result will thus be the equation
to the required locus.
Solving (3) and (4), we have
• /I ^^y
a Bin 6 =
and a cos 6 - c
7/ + (.r-6f'
_ y^ix-c)
~f+{^c^cf'
62 {x - c)
so that acos^ = c+ „ ,
y^ + {x- cy
Hence
a2 = a2 cos2 e + a^ sm^ e = c'^ + 2ch^ ^ ^~^ ^, + ^ .
y^+(x- c)- y^ + {x - c)2
The required locus is therefore
(a2 - c2) [?/2 + (.^ _ c)2] = 2c62 [x -c) + ¥.
This is a circle and its centre and radius are easily found.
154 COORDINATE GEOMETRY.
Ex. 3. Find the locus of a point P which is such that its polar with
respect to one circle touches a second circle.
Taking the notation of the last article, the equations to the two
circles are
x^ + if=a- (1),
and (a?-c)2 + ^2^62 (2).
Let {h, Jc) he the coordinates of any position of P. Its polar with
respect to (1) is
xh + yk = a^ (3),
Also any tangent to (2) has its equation of the form (Art. 179)
[x - c) cos d + y sin d = b (4).
If then (3) be a tangent to (2) it must be of the form (4).
^, „ cos d sin 9 c cos d + b
Therefore — - — = — ^ — = ,5 .
h k a"^
These equations give
cos 6 {a^-ch) = hh, and sin d{a^-ch) = hk.
Squaring and adding, we have
(a2-c7i)2=62(/i2+^2) (5).
The locus of the point {h, k) is therefore the curve
h'^{x^ + y^) = {a^-cxf.
Aliter. The condition that (3) may touch (2) may be otherwise
found.
For, as in Art. 153, the straight line (3) meets the circle (2) in the
points whose abscissae are given by the equation
k^{x-c)^ + {a^-hxf^b%\
i. e. x^- {h^ + F) - 2x {ck^ + a'^h) + [k^c"^ + a" - b'^k^) = 0.
The line (3) will therefore touch (2) if
(cZ;2 + a^hf={h^ + F) (Fc^ + a^- b^k%
i.e. a h^h^ + B) = {ch-aY-,
which is equation (5).
Ex. 4. is a fixed point and P any point on a given circle ; OP
is joined and on it a point Q is taken so that OP . OQ = a constant
quantity k- ; prove that the locus of Q is a circle which becomes a
straight line ivhen lies on the original circle.
THE CIRCLE. EXAMPLES. 155
Let O be taken as pole and the line through the centre C as the
initial line. Let OC = d, and let the
radius of the circle be a. Qyf\
The equation to the circle is then
a2 = 7-2 + d^ - 2rd cos d, (Art. 171), Q \d C"
where OP=r and lPOG = d.
Let OQ be p, so that, by the given
condition, we have rp-=k' and hence ?- = — .
P
Substituting this value in the equation to the circle, we have
a2=^VtZ2-2— cos^ (1),
so that the equation to the locus of Q is
' -2 55 2^cos^=- (2.
But the equation to a circle, whose radius is a' and whose centre is
on the initial line at a distance d', is
r2-2rd'cos0 = a'2-d'2 (3).
Comparing (1) and (2), we see that the required locus is a circle,
such that
Z'2/7 hi
,r- and n"^ d'^-
Hence a 2= __ -^, ^^^^^.^ - 1 J = ^-^ .
The required locus is therefore a circle, of radius ~z^ ^ , whose
d^ - a"
kH
centre is on the same line as the original centre at a distance — ^ ^
'^ d-- a^
from the fixed point.
When lies on the original circle the distance d is equal to a, and
the equation (1) becomes k'^ — 2drQ05dy i.e., in Cartesian coordinates,
_fc2
In this case the required locus is a straight line perpendicular
to OG.
When a second curve is obtained from a given curve by the above
geometrical process, the second curve is said to be the inverse of the
first curve and the fixed point O is called the centre of inversion.
The inverse of a circle is therefore a circle or a straight line
according as the centre of inversion is not, or is, on the circumference
of the original circle.
156 COORDINATE GEOMETRY.
Ex. 5. PQ is a straight line drawn through O, one of the common
points of two circles, and meets them'again in P and Q; find the locus of
the point S ivhich bisects the line PQ.
Take O as the origin, let the radii of the two circles be R and R',
and let the lines joining their centres to O make angles a and a' with
the initial line.
The equations to the two circles are therefore, {Art. 172 (2)},
r=2i?cos{^-a), and r = 2R' cos{d- a').
Hence, if S be the middle point of PQ, we have
20S =0P+0Q = 2R cos {d-a) + 2R' cos {6- a').
The locus of the point S is therefore
r = Rcos{d-a) + R'co3{e-a')
= {R cos a + R' cos a') cos ^ + (JR sin a + R' sin a'} sin 6
= 2R''co8{d-a") (1),
where 2R" cos a" = R cos a + R' cos a',
and 2R" sin a" =Rsma + R' sin a'.
Hence R" = i s/R^ + R'^ + 2RR' cos (a - a') ,
R sin a + R' sin a'
and tana"=
jR cos a + jR' cos a' '
From (1) the locus of S is a circle, whose radius is JR", which
passes through the origin and is such that the line joining O to its
centre is inclined at an angle a" to the initial line.
EXAMPLES. XXII.
1. A point moves so that the sum of the squares of its distances
from the four sides of a square is constant ; prove that it always lies
on a circle,
2. A point moves so that the sum of the squares of the perpendi-
culars let fall from it on the sides of an equilateral triangle is constant;
prove that its locus is a circle.
3. A point moves so that the sum of the squares of its distances
from the angular points of a triangle is constant ; prove that its locus
is a circle.
4. Find the locus of a point which moves so that the square of
the tangent drawn from it to the circle x^ + y^=a^ is equal to c times
its distance from the straight line lx + my + n = 0.
5. Find the locus of a point whose distance from a fixed point is
in a constant ratio to the tangent drawn from it to a given circle.
[EXS. XXII.] EXAMPLES. 157
6. Find the locus of the vertex of a triangle, given (1) its base and
the sum of the squares of its sides, (2) its base and the sum of m times
the square of one side and n times the square of the other.
7. A point moves so that the sum of the squares of its distances
from n fixed points is given. Prove that its locus is a circle.
8. Whatever be the value of a, prove that the locus of the inter-
section of the straight lines
X cos a + y sin a — a and x sin a-y cos a — b
is a circle.
9. From a point P on a circle perpendiculars P3I and FN are
drawn to two radii of the cu'cle which are not at right angles ; find
the locus of the middle point of 3IN.
10. Tangents are drawn to a circle from a point which always
lies on a given line ; prove that the locus of the middle point of the
chord of contact is another circle.
11. Find the locus of the middle points of chords of the circle
x^ + y^ = aP which pass through the fixed point {h, k).
12. Find the locus of the middle points of chords of the circle
x'^ + y^=a? which subtend a right angle at the point (c, 0).
13. is a fixed point and P any point on a fixed circle ; on OP
is taken a point Q such that OQ is in a constant ratio to OP ; prove
that the locus of Q is a circle.
14. is a fixed point and P any point on a given straight line ;
OP is joined and on it is taken a point Q such that OP . OQ = k-;
prove that the locus of Q, i. e. the inverse of the given straight line
with respect to 0, is a circle which passes through O.
15. One vertex of a triangle of given species is fixed, and another
moves along the circumference of a fixed circle ; prove that the locus
of the remaining vertex is a circle and find its radius.
16. O is any point in the plane of a circle, and OP^P^ any chord
of the circle which passes through O and meets the circle in Pj and
P. 2. On this chord is taken a point Q such that OQ is equal to (1) the
arithmetic, (2) the geometric, and (3) the harmonic mean between OP^
and 0P<^\ in each case find the equation to the locus of Q.
17. Find the locus of the point of intersection of the tangent to
any circle and the perpendicular let fall on this tangent from a fixed
point on the circle.
18. A circle touches the axis of x and cuts off a constant length
21 from the axis of y ; prove that the equation of the locus of its centre
is 2/^ - x^ = l^ cosec^ w, the axes being inclined at an angle w.
158 COORDINATE GEOMETHY. [ExS.
19. A straight line moves so that the product of the jDerpendi-
culars on it from two fixed points is constant. Prove that the locus
of the feet of the perpendiculars from each of these points upon the
straight line is a circle, the same for each.
20. is a fixed point and AP and BQ are two fixed parallel
straight lines ; BOA is perpendicular to both and POQ is a right
angle. Prove that the locus of the foot of the perpendicular drawn
from O upon PQ is the circle on AB as diameter.
21. Two rods, of lengths a and b, slide along the axes, which are
rectangular, in such a manner that their ends are always concyclic ;
prove that the locus of the centre of the circle passing through these
ends is the curve 4 {x^ - y'^) — a'^ - b^.
22. Shew that the locus of a point, which is such that the
tangents from it to two given concentric circles are inversely as the
radii, is a concentric circle, the square of whose radius is equal to the
sum of the squares of the radii of the given circles.
23. Shew that if the length of the tangent from a point P to the
circle x^ + y^=a-^ be four times the length of the tangent from it to the
circle {x - a)- + y^ = a?, then P lies on the circle
Ihx'^ + 15?/2 - ^lax + a2 = 0.
Prove also that these three circles pass through two points and that
the distance between the centres of the first and third circles is
sixteen times the distance between the centres of the second and
third circles.
24. Find the locus of the foot of the perpendicular let fall from
the origin upon any chord of the circle x- + y'^-\-2gx + 2fy + c=i(i which
subtends a right angle at the origin.
Find also the locus of the middle points of these chords.
25. Through a fixed point O are drawn two straight lines OPQ
and ORS to meet the circle in P and Q, and jR and S, respectively.
Prove that the locus of the point of intersection of PS and QR, as also
that of the point of intersection of PR and QS, is the polar of with
respect to the circle.
26. ^) -B, C, and D are four points in a straight line; prove that
the locus of a point P, such that the angles APB and CPD are equal,
is a circle.
27. The polar of P with respect to the circle x'^-\-y'^ — a- touches
the circle {x - of + {y - ^)"—b^ ; prove that its locus is the curve given
by the equation {ax + ^y-a^f — b^ {x^ + y^) .
28. A tangent is drawn to the circle {x - a)" + y" = b- and a perpen-
dicular tangent to the circle {x + a)'^ + y- = c" ; find the locus of their
point of intersection, and prove that the bisector of the angle between
them always touches one or other of two fixed circles.
XXII.] EXAMPLES. ' 159
29. Ill any circle prove that the perpendicular from any point of
it on the line joining the points of contact of two tangents is a mean
proportional between the perpendiculars from the point upon the two
tangents.
30. From any point on the circle
a;2 + if + 2gx + 2fy + c^0
tangents are drawn to the circle
x^ + ij'^ + 2gx + 2fy + c sin^ a + {g'^+f^) cos^ a = 0;
prove that the angle between them is 2a.
31. The angular points of a triangle are the points
(a cos a, a sin a), (acos/S, asin/3), and (a cos 7, a sin 7);
prove that the coordinates of the orthocentre of the triangle are
a (cos a + cos /3 + cos 7) and a (sin a + sin /3 + sin 7).
Hence prove that if ^i, B, C, and D be four points on a circle the
orthocentres of the four triangles ABC, BCD, CDA, and DAB lie on
a circle.
32. A variable circle passes through the point of intersection
of any two straight lines and cuts ofi from them portions OP and OQ
such that m . OP + n . OQ is equal to unity ; prove that this circle
always passes through a fixed point.
33. Find the length of the common chord of the circles, whose
equations are (x-af + y'^ = a? and x^ + (1/ - h)-=^JP, and prove that the
equation to the circle whose diameter is this common chord is
(a2 + &2) (a;2 + yi-^ ^2ab {bx + aij) .
34. Prove that the length of the common chord of the two circles
whose equations are
(a;-a)2 + (?/-fc)2 = c-2 and {x-hf + iy-af^c^
is V^c- - 2 (a - h)-K
Hence find the condition that the two circles may touch.
35. Find the length of the common chord of the circles
x^ + y^-2ax-4ay -4^0^ = and X' + y'-oax + 4:ay = 0.
Find also the equations of the common tangents and shew that
the length of each is 4a.
36. Find the equations to the common tangents of the circles
(1) x'^ + 7f-2x~&y + 9 = and x- + y-^ + ex-2y + 1 = 0,
(2) x^ + 7/2 = 6-2 and {x - a)^ + y^ = h-.
CHAPTER IX.
SYSTEMS OF CIRCLES.
[This chapter may he omitted hy the student on a first
reading of the subject.]
182. Orthogonal Circles.
said to intersect orthogonally when
the tangents at their points of
intersection are at right angles.
If the two circles intersect at
P, the radii O^P and O.^P., which
are perpendicular to the tangents
at P, must also be at right angles.
Def. Two circles are
Hence
O^Oi=O^P^-vO.JP\
i.e. the square of the distance between the centres must be
equal to the sum of the squares of the radii.
Also the tangent from Oo, to the other circle is equal to
the radius a^,, i.e. if two circles be orthogonal the length of
the tangent drawn from the centre of one circle to the
second circle is equal to the radius of the first.
Either of these two conditions will determine whether
the circles are orthogonal.
The centres of the circles
a;2 + 2/2 + 2f;^ + 2/?/ + c = and a;- + ?/2 + 2^'a; + 2/'y + c' = 0,
are the points {-g, -/) and (-^', -/') ; also the squares of their
radii are g^+f~- c and g'^ +f'^ - c'.
RADICAL AXIS OF TWO CIRCLES. 161
They therefore cut orthogonally if
i.e. if 2gg' + 2ff=c + c'.
1S3. Radical Axis. Def. The radical axis of
two circles is the locus of a point which moves so that the
lengths of the tangents drawn from it to the two circles are
equal.
Let the equations to the circles be
x' + y' + 2gx + 2fi/ + G^0 (1),
and x" + i/^ + 2gjX + 2fy + c^ = (2),
and let (x^, y^ be any point such that the tangents from it
to these circles are equal.
By Art. 168, we have
^1^ + Vx + ^9^1 + 2/2/1 + c = a^i" + 2/r + 2g^x^ + 2fy^ + c^,
i.e. 2x^{g-g,) + 2y,{f-f,) + G-C:, = 0.
But this is the condition that the point (iCj, 2/1) should
lie on the locus
2x{g-g,) + 2y{f-A) + c-c, = (3).
This is therefore the equation to the radical axis, and it
is clearly a straight line.
It is easily seen that the radical axis is perpendicular
to the line joining the centres of the circles. For these
centres are the points (— g, -f) and (-^1, — /i)- The
"m" of the line ioining them is therefore -^ — 7 {,
^ ^ -9x-{-9)
i.e. -tA.
9-9i
The "?7i" of the line (3) is - 7— §.
The product of these two " m's " is - 1 .
Hence, by Art. 69, the radical axis and the line joining
the centres are perpendicular.
L. 11
162
COORDINATE GEOMETRY.
184. A geometrical construction can be given
for the radical axis of two circles.
R,
^^^
\ >v
~~^^^/
^ ^v
/ \
A.
^ —
Ss
/ *
/ \
/^
7\
1
I
' \
o.
o
o.
Fig. 1.
Fig. 2.
If the circles intersect in real points, F and Q^ as in
Fig. 1, the radical axis is clearly the straight line PQ.
For if T be any point on PQ and TR and TS be the
tangents from it to the circles we have, by Euc. iii. 36,
TR'
TS\
TP . TQ
If they do not intersect in real points, as in the second
figure, let their radii be % and a^, and let 2^ be a point such
that the tangents TR and TS are equal in length.
Draw TO perpendicular to 0^0.2-
Since TR^=TS\
we have T0{- - O^R'' = TOi - O^S^
TO'' + 0^0'' - a^- - TO'' + OO.J - ai,
0^0''-00i=^a^-ai,
{0^0 - 00^) (0^0 + 00.) = ai' - a^\
0^0 — OOo =""7777^ = ^ constant quantity.
UiU.2
Hence is a fixed point, since it divides the fixed
straight line O^O^ into parts whose difference is constant.
Therefore, since O^OT is a right angle, the locus of T,
i.e. the radical axis, is a fixed straight line perpendicular to
the line joining the centres.
t.e.
%.e.
%.e,
I.e.
RADICAL AXIS. 163
185. If the equations to the circles in Art. 183 be
written in the form aS'=0 and aS" = 0, the equation (3) to
the radical axis may be written S — S' = 0, and therefore
the radical axis passes through the common points, real or
imaginary, of the circles S = and aS" = 0.
In the last article we saw that this was true geometri-
cally for the case in which the circles meet in real points.
When the circles do not geometrically intersect, as in
Fig. 2, we must then look upon the straight line TO as
passing through the imaginary points of intersection of the
two circles.
186. The radical axes of tJwee circles^ taken in 2)cdrs^
meet iri a j^oint.
Let the equations to the three circles be
^ = (1),
'S^'^O (2),
and S"=^0 (3).
The radical axis of the circles (1) and (2) is the straight
line
S->S' = (4).
The radical axis of (2) and (3) is the straight line
.S"-.S"'=:0 (5).
If we add equation (5) to equation (4) we shall have the
equation of a straight line through their points of inter-
section.
Hence .S'-aS'"-0 (6)
is a straight line through the intersection of (4) and (5).
But (6) is the radical axis of the circles (3) and (1).
Hence the three radical axes of the three circles, taken
in pairs, meet in a point.
This point is called the Radical Centre of the three
circles.
This may also be easily proved geometrically. For let
the three circles be called A, B, and C, and let the radical
axis of A and B and that of B and C meet in a point 0.
11—2
164 COORDINATE GEOMETKY.
By the definition of the radical axis, the tangent from
to the circle A = the tansrent from ^---^
to the circle B, and the tangent f V j
from to the circle B = tangent z'^' \ ^^\^^
from it to the circle C. f \ Ip
Hence the tangent from to >v ^/^v^' /^"^^
the circle A = the tangent from it ^^7\ \ \
to the circle C, i.e. is also a I ^^ j-
point on the radical axis of the V J
circles A and C. —
187. If S=0 and S' = he the equations of two circles,
the equation of any circle through their 2>oints of inter-
section is S — \S'. Also the equation to any circle, such that
the radical axis of it and S—0 is u = 0, is S + \u ~ 0.
For wherever S =0 and >S" = are both satisfied the
equation S = XS' is clearly satisfied, so that S = \S' is some
locus through the intersections of aS' = and *S"= 0.
Also in both S and S' the coefiicients of x^ and y'^ are
equal and the coefiicient of xy is zero. The same statement
is therefore true for the equation S=XS'. Hence the
proposition.
Again, since u is only of the first degree, therefore in
S + Xu the coefficients of ay^ and y^ are equal and the
coefficient of xy is zero, so that S + Xu = is clearly a circle.
Also it passes through the intersections of *S^ = and u — 0.
EXAMPLES. XXIII.
Prove that the following pairs of circles intersect orthogonally :
1. x'^ + y^-2ax + c = and x^ + y^ + 2by -c = 0.
2. x^ + y- - 2ax + 2hy + c = and x^ + y^ + 2hx + 2ay -c = 0.
3. Find the equation to the circle which passes through the origin
and cuts orthogonally each of the circles
x^ + y^-ex + 8 = and x^ + y^-2x-2y = 7.
Find the radical axis of the pairs of circles
4. ic2 + 2/2=144 and x^ + y^-15x + lly = 0.
5. x^ + y^-3x-4:y + 5=0 and Sx^- + By^-7x + 8y + ll=0.
RADICAL AXIS. EXAMPLES. 165
6. x^ + y--xy + Qx-ly + 8 = &nd x^ + ij' -xy-4 = 0,
the axes being inclined at 120°.
Find the radical centre of the sets of circles
7. x^+y^ + x + 2y + S = 0, x^ + 2f-h2x + 4:y + 5 = 0,
and x^ + y^-7x-8y-d = 0.
8. (a;-2)2+(z/-3)2 = 36, (re + 3)2 +(7/ + 2)2 = 49,
and (a;-4)2 + (i/ + 5)2=64.
9. Prove that the square of the tangent that can be drawn from
any point on one circle to another circle is equal to twice the product
of the perpendicular distance of the point from the radical axis of the
two circles, and the distance between their centres.
10. Prove that a common tangent to two circles is bisected by the
radical axis.
11. Find the general equation of aU circles any pair of which have
the same radical axis as the circles
j.2^y-2 — 4. an^ x^ + y^ + 2x + 4:y = &.
12. Find the equations to the straight lines joining the origin to
the points of intersection of
x^ + y^-4:x-2y = 4: and x'- + y^-2x~'iy -4^ = 0.
13. The polars of a point P with respect to two fixed circles meet
in the point Q. Prove that the circle on PQ as diameter passes
through two fixed points, and cuts both the given circles at right
angles.
14. Prove that the two circles, which pass through the two points
(0, a) and (0, - a) and touch the straight line y = mx + c, will cut ortho-
gonally if c2 = a2 (2 + m^).
15. Find the locus of the centre of the circle which cuts two given
circles orthogonally.
16. If two circles cut orthogonally, prove that the polar of any
point P on the first circle with respect to the second passes through
the other end of the diameter of the first circle which goes through P.
Hence, (by considering the orthogonal circle of three circles as
the locus of a point such that its polars with respect to the circles
meet in a point) prove that the orthogonal circle of three circles,
given by the general equation is
\x+9i> y+fi, 9i^+fiy+ci
\x + go, y + f^, g^x + f^y + c^ =0.
\x + gz, y + fs, g-i^+UJ+H
166 COORDINATE GEOMETEY.
188. Coaxal Circles. Def. A system of circles
is said to be coaxal when they have a common radical axis,
i.e. when the radical axis of each pair of circles of the
system is the same.
Tojind the equation of a system of coaxal circles.
Since, by Art. 183, the radical axis of any pair of the
circles is perpendicular to the line joining their centres, it
follows that the centres of all the circles of a coaxal system
must lie on a straight line which is perpendicular to the
radical axis.
Take the line of centres as the axis of x and the radical
axis as the axis of y (Figs. I. and II., Art. 190), so that
is the origin.
The equation to any circle with its centre on the axis
of £c is
x^ + y'^ — 2gx + c-0 (1).
Any point on the radical axis is (0, y^).
The square on the tangent from it to the circle (1) is,
by Art. 168, y^' + c.
Since this quantity is to be the same for all circles of
the system it follows that c is the same for all such circles ;
the different circles are therefore obtained by giving dif-
ferent values to g in the equation (1).
The intersections of (1) with the radical axis are then
obtained by putting a^ = in equation (1), and we have
If c be negative, we have two real points of intersection
as in Fig. I. of Art. 190. In such cases the circles are said
to be of the Intersecting Species.
If c be positive, we have two imaginary points of in-
tersection as in Fig. II.
'&•
189. Limiting points of a coaxal system.
The equation (1) of the previous article which gives any
circle of the system may be written in the form
(x-gY + 2/2 ^ / - c = [Jf - cf.
COAXAL CIRCLES.
167
It therefore represents a circle whose centre is the point
(^, 0) and whose radius is J g^ — c.
This radius vanishes, i.e. the circle becomes a point-
circle, when g^ — G, i.e. when g — ±Jc.
Hence at the particular points (+ Jc, 0) we have point-
circles which belong to the system. These point-circles are
called the Limiting Points of the system.
If G be negative, these points are imaginary.
But it was shown in the last article that when c is
negative the circles intersect in real points as in Fig. I.,
Art. 190.
If c be positive, the limiting points L^ and L^ (Fig. II.) are
real, and in this case the circles intersect in imaginary points.
The limiting points are therefore real or imaginary
according as the circles of the system intersect in imaginary
or real points.
190. Orthogonal circles of a coaxal system.
Let T be any point on the common radical axis
of a system of coaxal circles, and let TR be the tangent
from it to any circle of the system.
Then a circle, whose centre is T and whose radius is TR^
will cut each circle of the coaxal system orthogonally.
168
COORDINATE GEOMETRY.
[For the radius TR of this circle is at right angles to
the radius O^R, and so for its intersection with any other
circle of the system.]
Fig. II.
Hence the limiting points (being point- c^rcZes of the
system) are on this orthogonal circle.
The limiting points are therefore the intersections with
the line of centres of any circle whose centre is on the
common radical axis and whose radius is the tangent from
it to any of the circles of the system.
Since, in Eig. I,, the limiting points are imaginary these
orthogonal circles do not meet the line of centres in real
points.
In Fig. II. they jDass through the limiting points Z^
and 7^2 .
These orthogonal circles (since they all pass through two
points, real or imaginary) are therefore a coaxal system.
Also if the original circles, as in Fig. I., intersect in
real points, the orthogonal circles intersect in imaginary
points; in Fig. II. the original circles intersect in imaginary
points, and the orthogonal circles in real points.
We therefore have the following theorem :
A set of coaxal circles can be cut orthogonally hy another
set of coaxal circles, the centres of each set lying on the
radical axis of the other set ; also one set is of the limiting-
point sjyecies and the other set of the other species.
ORTHOGONAL CIRCLES. 169
191. Without reference to the limiting points of the original
system, it may be easily found whether or not the orthogonal circles
meet the original line of centres.
For the circle, whose centre is T and whose radius is TR, meets
or does not meet the line 0-fi,2 according as TR^ is > or < TO^,
i.e. according as TO^^-O^R^ is > TO^,
L e. according as TO^ +00^^- 0-fi^ is > TO^,
i.e. according as 00-^ is < O^R,
i.e. according as the radical axis is without, or within, each of the
circles of the original system.
192. In the next article the above results will be
proved analytically.
To find the equation to any circle y)hich cuts two ghien
circles orthogonally.
Take the radical axis of the two circles as the axis of y^
so that their equations may be written in the form
a? ^y^ — 2gx + c = (1),
and a? 4- y'^ — 2g^x +c — (2),
the quantity c being the same for each.
Let the equation to any circle which cuts them or-
thogonally be
{x-Ay + {y-Bf = E^ ....(3).
The equation (1) can be written in the form
{x-gf + f-^[J^::rcy (4).
The circles (3) and (4) cut orthogonally if the square of
the distance between their centres is equal to the sum of
the squares of their radii,
i.e. if {A - gf ^E^-^m^. [V/^]^
i.e. if A^-^B'-1Ag = R^-c (5).
Similarly, (3) will cut (2) orthogonally if
A--vB''-'lAg^ = E'-c (6).
Subtracting (6) from (5), we have A (g - g^) ^^^ 0.
Hence ^ = 0, and i?- = B^ + c.
170 COORDINATE GEOMETRY.
Substituting these values in (3), the equation to the
required orthogonal circle is
ar^ + 2/'-2%-c = (7),
where B is any quantity whatever.
Whatever be the value of B the equation (7) represents
a circle whose centre is on the axis of y and which passes
through the points (+ Jc, 0).
But the latter points are the limiting points of the
coaxal system to which the two circles belong. [Art. 189.]
Hence any pair of circles belonging to a coaxal system
is cut at right angles by any circle of another coaxal
system ; also the centres of the circles of the latter system
lie on the common radical axis of the original system, and
all the circles of the latter system pass through the limiting
points (real or imaginary) of the first system.
Also the centre of the circle (7) is the point (0, B) and
its radius is JB"^ + c.
. The square of the tangent drawn from (0, B) to the
circle {I) = B"" + c (by Art. 168).
Hence the radius of any circle of the second system is
equal to the length of the tangent drawn from its centre to
any circle of the first system.
193. The equation to the system of circles which cut
a given coaxal system orthogonally may also be obtained
by using the result of Art. 182.
For any circle of the coaxal system is, by Art. 188,
given by
a? + y'^- Igx + c = (1),
where c is the same for all circles.
Any point on the radical axis is (0, y).
The square on the tangent drawn from it to (1) is
therefore y"^ + c.
The equation to any circle cutting (1) orthogonally is
therefore
^^ + (2/ - 2/T = y"" + ^'
i.e. x^ + y'^— lyy —c = 0.
ORTHOGONAL CIRCLES. 171
Whatever be the value of y this circle passes through
the points (+ ^]c, 0), i.e. through the limiting points of the
system of circles given by (1).
194. We can now deduce an easy construction for the
circle that cuts any three circles orthogonally.
Consider the three circles in the figure of Art. 186.
By Art. 192 any circle cutting A and B orthogonally
has its centre on their common radical axis, i.e. on the
straight line OD.
Similarly any circle cutting B and C orthogonally has
its centre on the radical axis OE.
Any circle cutting all three circles orthogonally must
therefore have its centre at the intersection of OD and OE.,
i.e. at the radical centre 0. Also its radius must be the
length of the tangent dravi^n from the radical centre to
any one of the three circles.
Ex. Find the equation to the circle lohich cuts orthogonally each
of the three circles
x^ + 'ij^ + 2x + ny+ 4 = (1),
x'^ + 2/ + lx+ 6y + ll = (2),
x^ + if- x + 22y+ 3 = (3).
The radical axis of (1) and (2) is
5x-lly + 7 = 0.
The radical axis of (2) and (3) is
8x-l&y + 8 = 0.
These two straight Hnes meet in the point (3, 2) which is therefore
the radical centre.
The square of the length of the tangent from the point (3, 2) to
each of the given circles =57.
The required equation is therefore {x - 3)^ +{y - 2)^ = 57,
i. e. x^ + ?/2 — 6a; - 4?/ - 44 = 0.
195. Ex. Find the locus of a point which moves so that the length
of the tangent draion from it to one given circle is X times the length of
the tangent from it to another given circle.
As in Art. 188 take as axes of x and y the line joining the centres
of the two circles and the radical axis. The equations to the two
circles are therefore
x'^ + y''-2g^x + c = Q (1),
and a;2 + ^2_2^^a; + c = (2).
172 COORDINATE GEOMETRY.
Let (h, k) be a point such that the length of the tangent from it to
(1) is always X times the length of the tangent from it to (2).
Then Ji^ + k^- 2gji + c = \^ [/i^ + A;^ - ^.g^h + c].
Hence {h, h) always lies on the circle
x'^ + y^-<ix^-^^ + cr=0 (3).
A"' — 1
This circle is clearly a circle of the coaxal system to which (1) and
(2) belong.
Again, the centre of (1) is the point (f/^, 0), the centre of (2) is
(^2, 0), whilst the centre of (3) is ( ^toZ\^ ' ^) '
Hence, if these three centres be called O^ , 0.^ , and O3 , we have
"\ 2 1
and O^Os = "^^ _ { ^ ~ ^3 = -^YZi (^2 " ^i)>
so that O1O3 : 0^0^ : : X^ : 1.
The required locus is therefore a circle coaxal with the two given
circles and whose centre divides externally, in the ratio X" : 1, the line
joining the centres of the two given circles.
EXAMPLES. XXIV.
1. Prove that a common tangent to two circles of a coaxal
system subtends a right angle at either limiting point of the system.
2. Prove that the polar of a limiting point of a coaxal system
with respect to any circle of the system is the same for all circles of
the system.
3. Prove that the polars of any point with respect to a system of
coaxal circles all pass through a fixed point, and that the two points
are equidistant from the radical axis and subtend a right angle at a
limiting point of the system. If the first point be one limiting point
of the system prove that the second point is the other limiting point.
4. A fixed circle is cut by a series of circles all of which pass
through two given points ; prove that the straight line joining the
intersections of the fixed circle with any circle of the system always
passes through a fixed point.
5. Prove that tangents drawn from any point of a fixed circle of
a coaxal system to two other fixed circles of the system are in a
constant ratio.
[EXS. XXIV.] COAXAL CIRCLES. EXAMPLES. 173
6. Prove that a system of coaxal circles inverts with respect to
either limiting point into a system of concentric circles and find the
position of the common centre.
7. A straight line is drawn touching one of a system of coaxal
circles in P and catting another in Q and R. Shew that PQ and PR
subtend equal or supplementary angles at one of the limiting points
of the system.
8. Find the Ipcus of the point of contact of parallel tangents
which are drawn to each of a series of coaxal circles.
9. Prove that the circle of similitude of the two circles
x^- + y'^-2kx + b = and x- + if-2k'x + 5 = Q
{L e. the locus of the points at which the two circles subtend the same
angle) is the coaxal circle
10. From the preceding question shew that the centres of simili-
tude {i.e. the points in which the common tangents to two circles
meet the line of centres) divide the line joining the centres internally
and externally in the ratio of the radii.
11. If x + y sj -l = tan{u + v /sf -1), where x, y, u, and v are all
real, prove that the curves tt=: constant give a family of coaxal circles
passing through the points (0, ±1), and that the curves ?; = constant
give a system of circles cutting the first system orthogonally.
12. Find the equation to the circle which cuts orthogonally each
of the circles
x^-\-y'^ + '2.gx + c = 0, x'^-\-y^ + 2g'x-{-c = 0,
and x^ + y'^ + 2hx + 1ky + a=.0.
13. Find the equation to the circle cutting orthogonally the
three circles
x^ + y'^—a?', {x-cf-\-y'^=a'^, and x^ + {y -l))'" = a'^.
14. Find the equation to the circle cutting orthogonally the
three circles
and a;2 + 2/2 + 7^-9?/ + 29 = 0.
15. Shew that the equation to the circle cutting orthogonally the
circles
{x-aY + {y-hf = h\ {^x--bf+{y-af=d^
and {x-a-h-cY + y^^ah + c^,
is a;2 + 2/^-2a:(a + 6)-'?/(a + 6) + a2 + 3a6 + 62 = o.
CONIC SECTIONS.
CHAPTER X.
THE PARABOLA.
196. Conic Section. Def. The locus of a point
P, which moves so that its distance from a fixed point is
always in a constant ratio to its perpendicular distance
from a fixed straight line, is called a Conic Section.
The fixed point is called the Focus and is usually
denoted by S.
The constant ratio is called the Eccentricity and is
denoted by e.
The fixed straight line is called the Directrix.
The straight line passing through the Focus and per-
pendicular to the Directrix is called the Axis.
When the eccentricity e is equal to unity, the Conic
Section is called a Parabola.
When e is less than unity, it is called an SUipse.
When e is greater than unity, it is called a Hyper-
bola.
[The name Conic Section is derived from the fact that
these curves were first obtained by cutting a cone in
various ways.]
THE PARABOLA.
175
197. To find the equation to a Farahola.
Let S be the fixed point and ZM the directrix. We
require therefore the locus
of a point F which moves
so that its distance from S
is always equal to PM, its
perpendicular distance from
ZM.
Draw 8Z perpendicular
to the directrix and bisect
8Z in the point A ; produce
ZA to X,
The point A is clearly a
point on the curve and is
called the Vertex of the
Parabola.
Take A as origin, AX as the axis of x, and J.F,
perpendicular to it, as the axis of y.
Let the distance ZA^ or A8^ be called «, and let P be
any point on the curve whose coordinates are x and y.
Join /S'P, and draw PN and PM perpendicular respec-
tively to the axis and directrix.
We have then aSP^ = Pif ^
^.e. {x - of + 2/2 r= ZN'' = {a + xf,
y2 = 4ax (1).
This being the relation which exists between the co-
ordinates of any point P on the parabola is, by Art. 42, the
equation to the parabola.
Cor. The equation (1) is equivalent to the geometrical
proposition
PN^ = 4:AS.AIi.
198. The equation of the preceding article is the
simplest possible equation to the parabola. Throughout
this chapter this standard form of the equation is assumed
unless the contrary is stated.
176 COORDINATE GEOMETRY.
If instead of AX and A Y we take the axis and the
directrix ^M as the axes of coordinates, the equation
would be
(x — 2a)- + y^ = x",
i.e. y'^ = ia{x — a) (1).
Similarly, if the axis SX and a perpendicular line SL
be taken as the axes of coordinates, the equation is
aj2 + 2/2 = (a? + 2a)2,
i.e. y^ = 4:a(x + a) (2).
These two equations may be deduced from the equation
of the previous article by transforming the origin, firstly to
the point (- a, 0) and secondly to the point (a, 0).
199. The equation to the parabola referred to any focus and
directrix may be easily obtained. Thus the equation to the parabola,
whose focus is the point' (2, 3) and whose directrix is the straight
line X - 4y + S — 0, is
i. e. 17 [x^ + y^ - ix - 62/ + 13] :== [x^ + 16i/ + 9 - 8xy + Gx - 24.y] ,
i.e. 16x'^ + y^ + 8xy-7ix-78y + 212 = 0.
200. To trace the curve
2/2= iax (1).
If X be negative, the corresponding values of y are
imaginary (since the square root of a negative quantity is
unreal) ; hence there is no part of the curve to the left of
the point A.
If y be zero, so also is x, so that the axis of x meets
the curve at the point A only.
If X be zero, so also is y, so that the axis of y meets
the curve at the point A only.
For every positive value of x we see from (1), by taking
the square root, that y has two equal and opposite values.
Hence corresponding to any point P on the curve there
is another point P' on the other side of the axis which is
obtained by producing PN to P' so that PN and NP' are
THE PARABOLA. 177
equal in magnitude. The line PP' is called a double
ordinate.
As X increases in magnitude, so do the corresponding
values of y ; finally, when x becomes infinitely great, y
becomes infinitely great also.
By taking a large number of values of x and the
corresponding values of y it will be found that the curve is
as in the figure of Art. 197.
The two branches never meet but are of infinite length.
201. The quantity y"^ — 4aa;' is negative^ zero, or positive
according as the point {x\ y') is within, upon, or without the
parabola.
Let Q be the point {x , y') and let it be within the
curve, i.e. be between the curve and the axis AX. Draw
the ordinate QN and let it meet the curve in P.
Then (by Art. 197), PN^ - la . a;'.
Hence y"^, i.e. QN^, is < PiV^, and hence is < iax .
.'. y'^ — 4:ax is negative.
Similarly, if Q be without the curve, then y"-^, i.e. QJV%
is > PN^j and hence is > 4iax'.
Hence the proposition.
202. Latus Rectum. Def. The latus rectum of
any conic is the double ordinate LSL' drawn through the
focus S.
In the case of the parabola we have SL = distance of L
from the directrix — SZ= 2a.
Hence the latus rectum = la.
"When the latus rectum is given it follows that the
equation to the parabola is completely known in its
standard form, and the size and shape of the curve
determined.
The quantity la is also often called the principal
parameter of the curve.
Focal Distance of any point. The focal distance
of any point P is the distance 8P.
This focal distance = PM = ZN= ZA+AN'=a + x.
L. 12
178 COORDINATE GEOMETEY.
Ex. Find the vertex, axis, focus, and latus rectum of the parabola
4?/2 + 12ar-202/ + 67 = 0.
The equation can be written
y^-5y=-Sx--^^,
i.e. {y--^f=-Sx-s^- + ^^=-3{x + i).
Transform this equation to the point (-|, f) and it becomes
y^= -Sx, which represents a parabola, whose axis is the axis of x
and whose concavity is turned towards the negative end of this axis.
Also its latus rectum is 3.
Eef erred to the original axes the vertex is the point i-^, f ), the
axis is 2/ = f, and the focus is the point (-| -|, f), «-e. ( -V-i f)-
EXAMPLES. XXV.
Find the equation to the parabola with
1. focus (3, -4) and directrix Gcc- 7?/ + 5 = 0.
X XI
2. focus (a, &) and directrix - + f = 1.
^ ah
Find the vertex, axis, latus rectum, and focus of the parabolas
3. y^ = 4:X + ^y. 4. x'^ + 2y = 8x-7.
5. x^-2ax + 2ay = 0. 6. 2/^=4y-4a:.
7. Draw the curves
(1) y'2=-4:ax, (2) x'^=4:ay, and (3) x-z=-4:ay.
8 Find the value of p when the parabola y'^ = 4px goes through
the point (i) (3, - 2), and (ii) (9, - 12).
9. For what point of the parabola y^ = 18x is the ordinate equal
to three times the abscissa ?
10. Prove that the equation to the parabola, whose vertex and focus
are on the axis of x at distances a and a' f . om the origin respectively,
is y^ = 4:{a'-a){x-a).
11. In the parabola y^=Qx, find (1) the equation to the chord
through the vertex and the negative end of the latus rectum, and
(2) the equation to any chord through the point on the curve whose
abscissa is 24.
12. Prove that the equation y^ + 2Ax + 2By + C = represents a
parabola, whose axis is parallel to the axis of x, and find its vertex and
the equation to its latus rectum.
13. Prove that the locus of the middle points of all chords of
the parabola ?/2 = 4aa; which are drawn through the vertex is the
parabola y'^ = 2ax.
[EXS. XXV.] THE PARABOLA. EXAMPLES. 179
14. Prove that the locus of the centre of a circle, which intercepts
a chord of given length 2a on the axis of x and passes through a given
point on the axis of y distant 6 from the origin, is the curve
a;2-2'«/& + &2 = a2.
Trace this parabola.
15. PQ is a double ordinate of a parabola. Find the locus of its
point of trisection.
16. Prove that the locus of a point, which moves so that its
distance from a fixed line is equal to the length of the tangent drawn
from it to a given circle, is a parabola. Find the position of the
focus and directrix.
17. If a circle be drawn so as always to touch a given straight
line and also a given circle, prove that the locus of its centre is
a parabola.
18. The vertex ^ of a parabola is joined to any point P on the
curve and PQ is drawn at right angles to AP to meet the axis in Q.
Prove that the projection of PQ on the axis is always equal to the
latus rectum.
19. If on a given base triangles be described such that the sum of
the tangents of the base angles is constant, prove that the locus of
the vertices is a parabola.
20. A double ordinate of the curve y^=^px is of length 8p ; prove
that the lines from the vertex to its two ends are at right angles.
21. Two parabolas have a common axis and concavities in oppo-
site directions ; if any line parallel to the common axis meet the
parabolas in P and P', prove that the locus of the middle point of PP'
is another parabola, provided that the latera recta of the given para-
bolas are unequal.
22. A parabola is drawn to pass through A and P, the ends of
a diameter of a given circle of radius a, and to have as directrix a
tangent to a concentric circle of radius h ; the axes being AB and
a perpendicular diameter, prove that the locus of the focus of the
parabola IS - + ,^^=1,
203. To find the points of intersection of any straight
line with the parabola
2/^ = 4acc (1).
The equation to any straight line is
y = 7nx + c .(2).
The coordinates of the points common to the straight
line and the parabola satisfy both equations (1) and (2),
and are therefore found by solving them.
12—2
180 COORDINATE GEOMETRY.
Substituting the value of y from (2) in (1), we have
{mx + cf — 4:ax,
i.e. m^oc^ + 2x (mc - 2a) + G^ - (3).
This is a quadratic equation for x and therefore has two
roots, real, coincident, or imaginary.
The straight line therefore meets the parabola in two
points, real, coincident, or imaginary.
The roots of (3) are real or imaginary according as
{2(mc-2a)f-47^iV
is positive or negative, i.e. according as — aTnc + a^ is
positive or negative, i.e. according as mc is ^ ia.
204. To find the length of the chord intercepted by the parabola on
the straight line
y — mx + c (1).
If (o^i , y-^ and {x^, y^) he the common points of intersection, then,
as in Art. 154, we have, from equation (3) of the last article,
{^x^ — x^j = (ajj + x^) — ^x^x^
_4(mc-2a)^ 4:C^ _l&a{a-mc)
~ m* m^ wi'* '
and yj^-y^ = m{x^-x^).
Hence the required length = \/(l/i - 2/2)^ + (^1 ~ ^"2)^
= Jl + m'-^ (x^ - ajg) = ^2 Jl + mP J a (a - mc).
205. To find the equation to the tangent at any point
(^'3 y) 9f i^^^ parabola y^ = 4:ax.
The definition of the tangent is given in Art. 149.
Let P be the point [x, y) and Q a point (x", y") on the
parabola.
The equation to the line PQ is
2/-y' = f5f'(^-»'') (!)•
Since P and Q both lie on the curve, we have
y'^^4ax' (2),
and y"^^4.aa/' (3).
TANGENT AT ANY POINT OF A PARABOLA. 181
Hence, by subtraction, we have
2/"' - y"' ^ 4a (x" - x),
i.e.
{y"-y'){y" + y') = 4.a{x"-x'),
and hence
y" -y 4o&
X —X y -^ y
Substituting this value in equation (1), we have, as
the equation to any secant PQ^
4a
i.e. y {y + y") = ^ax + y'y" + y^ — 4:ax
— iax + y'y" (4).
To obtain the equation of the tangent at (x\ y') we take
Q indefinitely close to P, and hence, in the limit, put y" = y .
The equation (4) then becomes
^yy — y"^ + 4aa; = 4^ax + 4aa;',
i.e. yy' = 2a(x + x').
Cor. It will be noted that the equation to the tangent
is obtained from the equation to the curve by the rule of
Art. 152.
Exs. The equation to the tangent at the point {2, -4) of the
parabola y^=Qx is
2/(-4) = 4(a; + 2),
i.e. x-\-y + 1 = 0.
The equation to the tangent at the point ( — 2 » — ) of the parabola
y^=4iax is
2a . f a\
y . — = 2a[x + ~] ,
d
I.e. ii = mx-\ — .
m
206. To find the condition that the straight line
y = mx + c (1)
rtiay touch the parabola y^ — 4:ax (2).
The abscisses of the points in which the straight line (1)
meets the curve (2) are as in Art. 203, given by the equation
rn^x^ + 2x {mo - 2a) + c^ = (3).
182 COORDINATE GEOMETRY.
The line (1) will touch (2) if it meet it in two points
which are indefinitely close to one another, i.e. in two
points which ultimately coincide.
The roots of equation (3) must therefore be equal.
The condition for this is
4 {mc — 2ay — iirrc^
i.e. ct? - amc — 0,
a
so that G = — .
m
Substituting this value of c in (1), we have as the
equation to a tangent,
a
y = mx + — .
m
In this equation tyi is the tangent of th6 angle which
the tangent makes with the axis of x.
The foregoing proposition may also be obtained from the equation
of Art. 205.
For equation (4) of that article may be written
la lax' ,^.
y=—x+^ (1).
y y
In this equation put — , =m, i.e. y =— ,
^ y m
, y'^ a , 2ax' a
and hence x = -r- = ^, , and — — = — .
4a m^ y m
a
The equation (1) then becomes y = mx-] — .
Also it is the tangent at the point {x', y'), i.e. (—^, — j .
207. Equation to the normal at (x, y'). The required
normal is the straight line which passes through the point
{x\ y') and is perpendicular to the tangent, i.e. to the
straight line
2a . ,.
yz=z—-{x + x).
Its equation is therefore
y — y = rri' {x — x'),
2fi 11
where m' x — rr - 1 i.e. ^i' = -^ (Art. 69.)
y 2a
NORMAL TO A PARABOLA.
183
,(1).
and the equation to the normal is
y-y'=^(x-x')
208. To exj^ress the equation of the normal in the form
y = mx — 2a7n — am^.
In equation (1) of the last article put
2a
= m, %.e. y
- — ^.
am.
/2
Hence
, y 2
X =^ = aiir.
4a
The normal is therefore
y + 2am = m {x
am
%
t.e.
y = mx — 2am — am^
and it is a normal at the point (am^, — 2am) of the curve.
In this equation m is the tangent of the angle which
the normal makes with the axis. It must be carefully
distinguished from the m of Art. 206 which is the tangent
of the angle which the tangent makes with the axis. The
" m" of this article is - 1 divided by the " m'' of Art. 206.
209. Subtangent and Subnormal. Def. If
the tangent and normal at any point P of a conic section
meet the axis in T and G respectively and PN be the
ordinate at P, then NT is called the Subtangent and NG the
Subnormal of P.
To find the length of the subtangent and suhnorm^al.
If P be the point {x\ y') the equation to TP is, by
Art. 205,
yy —2a{x-\- x) (1).
To obtain the length of AT^ we
have to find the point where this
straight line meets the axis of tc,
i.e. we put 2/ = 0in (1) and we
have
x=^-x (2).
Hence AT=AN,
184 COORDINATE GEOMETRY.
[The negative sign in equation (2) shews that T and
N always lie on opposite sides of the vertex -4.]
Hence the subtangent iV^= 2J^iV = twice the abscissa
of the point P.
Since TFG is a right-angled triangle, we have (Euc. vi. 8)
FN'^^TN.NG.
Hence the subnormal NG
_ PiP _ PN^
The subnormal is therefore constant for all points on
the parabola and is equal to the semi-latus rectum.
210. Ex. 1. If a chord which is normal to the parabola at one
end subtend a right angle at the vertex, prove that it is inclined at an
angle tan~^ ^J2 to the axis.
The equation to any chord which is normal is
y = mx — 2am - am^,
i.e. mx-y = 2am+am^.
The parabola is y^ — 4:ax.
The straight lines joining the origin to the intersections of these
two are therefore given by the equation
y^ {2am + am^) - iax {mx -y) = 0.
If these be at right angles, then
2am + am^ — 'iam = 0,
i.e. m= ^sJ2.
Ex. 2. From the point where any normal to the parabola y^ = ^ax
meets the axis is draion a line perpendicular to this normal ; prove that
this line always touches an equal parabola.
The equation of any normal to the parabola is
y = mx — 2am - am^.
This meets the axis in the point {2a + am'^, 0).
The equation to the straight line through this point perpendicular
to the normal is
y = wii {x-2a — am'^) ,
where m^m= - 1.
The equation is therefore
y = m,[x-2a-^^,
i.e. y = m^{x-2a) .
TANGENT AND NORMAL. EXAMPLES. 185
This straight line, as in Art. 206, always touches the equal parabola
y^= - 4a (a;- 2a),
whose vertex is the point (2a, 0) and whose concavity is towards the
negative end of the axis of x.
EXAMPLES. XXVI.
Write down the equations to the tangent and normal
1. at the point (4, 6) of the parabola y^=9x,
2. at the point of the parabola ?/^ = 6a; whose ordinate is 12,
3. at the ends of the latus rectum of the parabola y^ — 12x,
4. at the ends of the latus rectum of the parabola ^2 — 4.^ (.^ _ a).
5. Find the equation to that tangent to the parabola y^ = 7x
which is parallel to the straight line 4y -x + S = 0. Find also its
point of contact.
6. A tangent to the parabola y^=Aax makes an angle of 60° with
the axis ; find its point of contact.
7. A tangent to the parabola y'^ = 8x makes an angle of 45° with
the straight line y = Sx + 5. Find its equation and its point of
contact.
8. Find the points of the parabola y^ = 4:ax at which (i) the
tangent, and (ii) the normal is inclined at 30° to the axis.
9. Find the equation to the tangents to the parabola y^=9x which
goes through the point (4, 10).
10. Prove that the straight line x + y = l touches the parabola
y=x-x^.
11. Prove that the straight line y = mx + c touches the parabola
?/^=4a (a; + a) if c=ma + —.
' m
12. Prove that the straight line Ix + my + w = touches the parabola
y^=4:ax if ln=amP.
13. For what point of the parabola y^ = 4:ax is (1) the normal equal
to twice the subtangent, (2) the normal equal to the difference between
the subtangent and the subnormal ?
Find the equations to the common tangents of
14. the parabolas ?/2 = 4aa; and .'r2 = 4&i/,
15. the circle x^ + y^=4:ax and the parabola y^=4:ax.
16. Two equal parabolas have the same vertex and their axes are
at right angles ; prove that the common tangent touches each at the
end of a latus rectum.
186 COOKDINATE GEOMETRY. [ExS.
17. Prove that two tangents to the parabolas y^ — 4a {x + a) and
y^=4:a' {x + a'), which are at right angles to one another, meet on the
straight line x + a + a' = 0.
Shew also that this straight line is the common chord of the two
parabolas.
18. PN is an ordinate of the parabola ; a straight line is drawn
parallel to the axis to bisect NP and meets the curve in Q ; prove
that NQ meets the tangent at the vertex in a point T such that
AT = %NP.
19. Prove that the chord of the parabola y^ — 'iax, whose equation
isy -'XiJ2 + 4:a^2 = 0, is a normal to the curve and that its length is
6 ^Sa.
20. If perpendiculars be drawn on any tangent to a parabola from
two fixed points on the axis, which are equidistant from the focus,
prove that the difference of their squares is constant.
21. If P, Q, and R be three points on a parabola whose ordinates
are in geometrical progression, prove that the tangents at P and R
meet on the ordinate of Q.
22. Tangents are drawn to a parabola at points whose abscissae
are in the ratio fi : 1; prove that they intersect on the curve .
y^={fi^ + fi~^)^ax.
23. If the tangents at the points {x', y') and {x", y") meet at the
point [x-^, y-j) and the normals at the same points in {x^, y^, prove
that
(1) .,=y^ .ni y,=y^f ,
(2) .,=2a + ^'^±^-;^^ and V.^-yV^^,
and hence that
(3) x,=2a+y-^- X, and y, = - "^^^^ .
24. From the preceding question prove that, if tangents be drawn
to the parabola y^ = 4:ax from any point on the parabola y^ — a{x+h),
then the normals at the points of contact meet on a fixed straight
line.
25. Find the lengths of the normals drawn from the point on the
axis of the parabola y^ = 8ax whose distance from the focus is 8a.
26. Prove that the locus of the middle point of the portion of a
normal intersected between the curve and the axis is a parabola whose
vertex is the focus and whose latus rectum is one quarter of that of
the original parabola.
27. Prove that the distance between a tangent to the parabola and
the parallel normal is a cosec 6 sec^ 6, where 6 is the angle that either
makes with the axis.
XXVI.l TANGENT AND NORMAL. EXAMPLES. 187
28. PNP' is a double ordinate of the parabola ; prove that the
locus of the point of intersection of the normal at P and the diameter
through P' is the equal parabola y^ = 4a (x-Aa).
29. The normal at any point P meets the axis in G and the
tangent at the vertex in G' ; HA be the vertex and the rectangle
AGQG' he completed, prove that the equation to the locus of Q is
30. Two equal parabolas have the same focus and their axes are
at right angles ; a normal to one is perpendicular to a normal to the
other ; prove that the locus of the point of intersection of these
normals is another parabola.
31. If a normal to a parabola make an angle with the axis,
shew that it will cut the curve again at an angle tan~i (^ tan 0).
32. Prove that the two parabolas y^ = 4:ax and y^=4:c{x- b) cannot
have a common normal, other than the axis, unless >2.
a-c
33. If aP>8h-, prove that a point can be found such that the two
tangents from it to the parabola y^=4tax are normals to the parabola
x^=^by.
34. Prove that three tangents to a parabola, which are such that
the tangents of their inclinations to the axis are in a given harmonical
progression, form a triangle whose area is constant.
35. Prove that the parabolas y^=4tax and x^ = 4:by cut one another
at an angle tan ^
2 {a« + 6«)
36. Prove that two parabolas, having the same focus and their axes
in opposite directions, cut at right angles.
37. Shew that the two parabolas
x^ + 4:a{y-2b-a) = and y'^ = 4:b{x-2a + b)
intersect at right angles at a common end of the latus rectum
of each.
38. ^ parabola is drawn touching the axis of x at the origin and
having its vertex at a given distance k from this axis. Prove that the
axis of the parabola is a tangent to the parabola x'^= -Sk {y -2k).
211. Some properties of the Parabola.
(a) If the tangent and normal at any point P of the
parabola meet the axis in T and G respectively, then
188
COORDINATE GEOMETRY.
and the tangent at P is equally inclined to the axis and the
focal distance of P.
Let P be the point (x, y).
Draw PM perpendicular to the directrix.
By Art. 209, we have AT^AN.
:. TS=TA + AS=^AF+ZA = ZF=MP = SP,
and hence z STP = z SPT.
By the same article, NG - "iAS = ZS.
:. SG^SN-\-NG = ZS+SF=MP = SP.
(/8) If the tangent at P meet the directrix in K, then
KSP is a right angle.
Por z SPT=^ L PTS=^ L KPM.
Hence the two triangles KPS and KPM have the two
sides KPj PS and the angle KPS equal respectively to the
two sides KP, PM and the angle KPM.
Hence z KSP = z KMP = a right angle.
Also lSKP=lMKP.
(y) Tangents at the extremities of any focal chord inter-
sect at right angles in the directrix.
For, if PS be produced to meet the curve in P', then,
since z P'SK is a right angle, the tangent at P' meets the
directrix in K
PROPERTIES OF THE PARABOLA. 189
Also, by (13), L MKP = z SEP,
and, similarly, / M'KP' - L SKF.
Hence
z PKP' = J z: SKM + 1 z SKM' = a right angle.
(8) 7/ /Sl^ 6e jyerpe^Lclicular to the tangent at P, then Y
lies on the tangent at the vertex and SY^ = AS . SP.
For the equation to any tangent is
y—mx-\ — (Ij.
The equation to the perpendicular to, this line passing
through the focus is
2/ = --(^-«) (2).
The lines (1) and (2) meet where
a \ , . 1 a
nix H — =— —[x— a) = X -^ — 5
7n m m in
i. e. where x — 0.
Hence Y lies on the tangent at the vertex.
Also, by Euc. vi. 8, Cor.,
SY^ = SA.ST=AS.SP,
212. To prove that through any given point {x^^ y^
there pass, in general, two tangents to the parabola.
The equation to any tangent is (by Art. 206)
y = mx -\ — ( 1 ).
If this pass through the fixed point (x^, y^), we have
a
y, = TUX, + — ,
i. e. m^Xj^ — tny^ + ^ = (2).
For any given values of x^ and y^ this equation is in
general a quadratic equation and gives two values of m
(real or imaginary).
Corresponding to each value of in we have, by substi-
tuting in (1), a different tangent.
190 COORDINATE GEOMETRY.
The roots of (2) are real and different if y-^ — 4:ax-^ be
positive, i.e., by Art. 201, if the point {x-^, y-^) lie without
the curve.
They are equal, i. e. the two tangents coalesce into one
tangent, if yi— ^cix^ be zero, i.e. if the point {x-^, y^ lie on
the curve.
The two roots are imaginary if y^ — 4a.x\ be negative,
i.e. if the point (cCj, y^ lie within the curve.
213. Equation to the chord of contact of tangents
drawn from a point {x^, y^).
The equation to the tangent at any point Q, whose
coordinates are x' and y', is
yy' = 2a (x + x).
Also the tangent at the point E, whose coordinates are
x" and y", is
yy" — 2a{x + x").
If these tangents meet at the point T, whose coordi-
nates are x^ and y^, we have
y^y' = 2a{x^+ x) (1)
and y^y" = 2a{x^ + x") (2).
The equation to QR is then
3ryi = 2a(x + Xi) (3).
For, since (1) is true, the point {x, y') lies on (3).
Also, since (2) is true, the point {x", y") lies on (3).
Hence (3) must be the equation to the straight line
joining ix\ y) to the point {x' , y"), i. e. it must be the
equation to QR the chord of contact of tangents from the
point {x^, ?/i).
214. The polar of any point with respect to a para-
bola is defined as in Art. 162.
To find the equation of the polar of the point [x^ , 2/1)
with respect to the parabola y^ — ^ax.
Let Q and R be the points in which any chord drawn
through the point P, whose coordinates are (x^, y^), meets
the parabola.
THE PARABOLA. POLE AND POLAR.
191
Let the tangents at Q and R meet in the point whose
coordinates are (A, k).
T(h.Wji.
We require the locus of (h, k).
Since ^^ is the chord of contact of tangents from (7i, k)
its equation (Art. 213) is
ky = 2a(x + h).
Since this straight line passes through the point (r^ , y^)
we haye
%i = 2a{x^ + h) (1).
Since the relation (1) is true, it follows that the point
{hj k) always lies on the straight line
3ryi = 2a(x + xJ (2).
Hence (2) is the equation to the polar of (ic^, y^.
Cor. The equation to the polar of the focus, viz. the point [a, 0),
is Q = x + a, so that the polar of the focus is the directrix.
215. When the point (x-^,y^ lies without the parabola
the equation to its polar is the same as the equation to the
chord of contact of tangents drawn from [x-^^, y^).
When (x^, y^) is on the parabola the polar is the same
as the tangent at the point.
As in Art. 164 the polar of (a^, y^) might have been
defined as the chord of contact of the tangents (real or
imaginary) that can be drawn from it to the parabola.
216. Geometrical construction for the polar of a point
192
COORDINATE GEOMETRY.
Let T be the point {x^^ 2/1)3 so that its polar is
yy^=-2a{x + x^) (1).
Through T draw a straight line parallel to the axis ; its
equation is therefore
y=yi (2).
Let this straight line meet the polar
in V and the curve in P.
The coordinates of F, which is the
intersection of (1) and (2), are therefore
^ —x^ and 2/1 (3).
Also P is the point on the curve
whose ordinate is y^, and whose coordi-
nates are therefore
2
and 2/1.
yi
4:a
Since abscissa of P=
abscissa of :Z^ + abscissa of V
there-
fore, by Art. 22, Cor., P
middle point of TV.
Also the tangent at P is
2/1'
is
the
yy,= 2a^.^f^
which is parallel to (1).
Hence the polar of T is parallel
to the tangent at P.
To draw the polar of T we therefore draw a line through
T, parallel to the axis, to meet the curve in P and produce
it to Fso that TP-PV; a line through F parallel to the
tangent at P is then the polar required.
217. If the polar of a point P passes through the point T, then
the polar of T goes through P. (Fig. Art. 214).
Let P be the point (x^, y-^) and T the point {h, k).
The polar of P is yy^ = 2a{x + x^).
Since it passes through T, we have
yj^k = 2a{x-^ + h) (1).
PAIR OF TANGENTS FROM ANY POINT. 193
The polar of T isyk = 2a (x+h).
Since (1) is true, this equatio n is satisfied by the coordinates Xj^
and t/i-
Hence the proposition.
Cor. The point of intersection, T, of the polar s of two points,
P and Q, is the pole of the line PQ.
218. To find the pole of a given straight line ivith respect to the
parabola.
Let the given straight line be
Ax + By+C=0.
If its pole be the point {x^, y-^), it must be the same straight
line as
yy^ = 2a{x + x^),
i.e. 2ax - yyi + 2axj^ = 0.
Since these straight lines are the same, we have
2a _ -yi _ 2axi
G ^ 2Ba
I.e. xi = j and y^= - -j- -
219. To find the equation to the pcdr of tangents that
can he drawn to the parabola from the point {x^^ y^.
Let (A, k) be any point on either of the tangents drawn
from (rL'i, y^. The equation to the line joining (x^, y^) to
(^, k) is
k— y. hy, - kx.
%.e. y = - — -x^-^ \
If this be a tangent it must be of the form
a
y — mx -{ — ,
, , . k — y^ , hy. — kx. a
so that . — ^ = m and ~- i = — .
a — x^ h — x^ m
Hence, by multiplication,
k — y^ hy^ — kx^
i. e. a (lb - x^^ = {k — y^) [hy^ — kx^.
I^ 13
194 COOKDINATE GEOMETRY.
The locus of the point (A, k) {i. e. the pair of tangents
required) is therefore
a(x-x^y = {y-y^) {xy^-yx:^ (1).
It will be seen that this equation is the same as
{f - \ax) (2/1^ - 4arci) = {2/2/1 - 2« (a? + x^f.
220. To prove that the middle points of a system of
parallel chords of a parabola all lie on a straight line which
is parallel to the axis.
Since the chords are all parallel, they all make the same
angle with the axis of x. Let Q
the tangent of this angle be on.
The equation to QB, any-
one of these chords, is there- y^^-
fore
y - mx + c (1 ), '^,
where c is different for the
several chords, but 7n is the
same.
This straight line meets the parabola y^ = 4:ax in points
whose ordinates are given by
m,y^ = 4:a (y — c),
4:a Aac , .
I.e. V y + =^0 (2).
^ m ^ m • '
Let the roots of this equation, i.e. the ordinates of Q
and Rj be y' and y'\ and let the coordinates of F, the
middle point of QR, be (h, k).
Then, by Art. 22,
T _ y + y" _ 2«
2 m
from equation (2).
The coordinates of V therefore satisfy the equation
2a
y=m^
so that the locus of F is a straight line parallel to the axis
of the curve,
MIDDLE POINTS OF PARALLEL CHORDS. 195
2a
The straight line 3/ = — meets the curve in a point P,
whose ordinate is — and whose abscissa is therefore — x .
m m"
The tangent at this point is, by Art. 205,
a
y = Tnx -\ — ,
and is therefore parallel to each of the given chords.
Hence the locus of the middle points of a system of
parallel chords of a parabola is a straight line which is
parallel to the axis and meets the curve at a point the
tangent at which is parallel to the given system.
221. To find the equation to the chord of the parabola ivhich is
bisected at any point {h, Jc).
By the last article the required chord is parallel to the tangent at
the point P where a line through {h, k) parallel to the axis meets the
curve.
Also, by Art. 216, the polar of {h, k) is parallel to the tangent at
this same point P.
The required chord is therefore parallel to the polar yJc = 2a {x + h).
Hence, since it goes through {h, k), its equation is
k{y-k) = 2a{x- h) (Art. 67).
222. Diameter. Def. The locus of the middle points
of a system of parallel chords of a parabola is called a
diameter and the chords are called its ordinates.
Thus, in the figure of Art. 220, PF is a diameter and
QB and all the parallel chords are ordinates to this
diameter.
The proposition of that article may therefore be stated
as follows.
Any diameter of a parabola is parallel to the axis and
the tangent at the point where it Tneets the curve is parallel
to its ordinates.
223. The tangents at the ends of any chord meet on
the diameter which bisects the chord.
Let the equation of QR (Fig., Art. 220) be
y = mx + c (1),
13—2
196
COORDINATE GEOMETRY.
and let the tangents at Q and R meet at the point T
Then QR is the chord of contact of tangents drawn
from T^ and hence its equation is
2/2/1 = 2a{x + x^) (Art. 213).
Comparing this with equation (1), we have
2a ,, , 2a
— = m, so that Vi = — ->
2/1 ^*
and therefore T lies on the straight line
2a
^ m
But this straight line was proved, in Art. 220, to be
the diameter P V which bisects the chord.
224. To find the equation to a parabola, the axes
being any diameter and the tangent to the parabola at the
point where this diameter meets the curve.
Let PVX be the diameter and PY the tangent at P
meeting the axis in T.
Take any point Q on the curve,
and draw QM perpendicular to the
axis meeting the diameter P F in L.
Let PVhQ X and VQ be y.
Draw PN perpendicular to the
axis of the curve, and let
e^ /. YPX=iPTM,
Then
iAS. A]S[^PN^ = ]SfT^ ts,Ti^e=^.AN^ . tan^ 6.
:. ANr=:AS. cot^ e = a cot^ e,
and PN = JIASTaN = 2a cot 6.
Now QM'- = 4:AS.AM=4:a.AM (1).
Also
QM=JSrP + LQ = 2acote+ VQsmO = 2acotO+ysinO,
and AM=A]\/' + PV+ VL=-acot^e + x + ycose.
THE PARABOLA. EXAMPLES. 197
Substituting these values in (1), we have
(2a cot + y sin Oy — ia (a cot^ + x + y cos 6),
i. e. if- sin^ 6 — ^ax.
The required equation is therefore
y'^^lpx (2),
where
p - T^= « (1 + ^ot' Q) = a^ AN= SP (by Art. 202).
The equation to the parabola referred to the above axes
is therefore of the same form as its equation referred to the
rectangular axes of Art. 197.
The equation (2) states that
QV'^^iSP.PV.
225. The quantity 4^j is called the parameter of the
diameter P V. It is equal in length to the chord which is
parallel to P F and passes through the focus.
For if Q'V'R' be the chord, parallel to PZand passing
through the focus and meeting PT in V\ we have
PY' = ST=SP^p,
so that Q' V"" ^ip.PV'^ ip\
and hence Q'R' =-'2Q'V' ^ ip.
226. Just as in Art. 205 it could now be shown that
the tangent at any point {x\ y) of the above curve is
yy — 2p (x + x).
Similarly for the equation to the polar of any point.
EXAMPLES. XXVII.
1. Prove that the length of the chord joining the points of
contact of tangents drawn from the point (Xj, y^ is
ijy-^ + 4a2 fjy^^ - 4aa; J ^
a
2. Prove that the area of the triangle formed by the tangents
3
from the point {x^^ y^ and the chord of contact is {y^ - ^ax^^ -^2a.
198 COORDINATE GEOMETRY. [Exs. XXVII.]
3. If a perpendicular be let fall from any point P upon its polar
prove that the distance of the foot of this perpendicular from the
focus is equal to the distance of the point P from the directrix.
4. What is the equation to the chord of the parabola y^ = 8x
which is bisected at the point (2, - 3) ?
5. The general equation to a system of parallel chords in the
parabola y^ = ^x is 4:X-y + k = 0.
Wliat is the equation to the corresponding diameter ?
6. P, Q, and B are three points on a parabola and the chord PQ
cuts the diameter through R in V. Ordinates P3I and QN are drawn
to this diameter. Prove that RM . RN=RV^.
7. Two equal parabolas with axes in opposite directions touch at
a point O. From a point P on one of them are drawn tangents PQ
and PQ' to the other. Prove that QQ' will touch the first parabola in
■P' where PP' is parallel to the common tangent at O.
Coordinates of any point on the parabola ex-
pressed in terms of one variable.
227. It is often convenient to express the coordinates
of any point on the curve in terms of one variable.
It is clear that the values
a 2a
mr 7n
always satisfy the equation to the curve.
Hence, for all values of m, the point
a 2a\
lies on the curve. By Art. 206, this m is equal to the
tangent of the angle v^hich the tangent at the point makes
v^^ith the axis.
The equation to the tangent at this point is
a
y = nix -\ — ,
and the normal is, by Art. 207, found to be
a
my + X = 2a + —-, .
in-
COORDINATES IN TERMS OF ONE VARIABLE. 199
228. The coordinates of the point could also be ex-
pressed in terms of the m of the normal at the point ; in
this case its coordinates are am?' and — 2am,
The equation of the tangent at the point (am^, — 2a7n)
is, by Art. 205,
mi/ + X + am^ — 0,
and the equation to the normal is
y — mx — 1am, — am?.
229. The simplest substitution (avoiding both nega-
tive signs and fractions) is
X = at2 and y = 2at.
These values satisfy the equation y^ = ^ax.
The equations to the tangent and normal at the point
{af, 2at) are, by Arts. 205 and 207,
ty = x + at^,
and y + tx= 2at + af.
The equation to the straight line joining
(atj^, 2at-^ and {at^, 2at^
is easily found to be
y {h + ^2) = 2x + 2at-f^.
The tangents at the points
{at^^ 2at^ and iat^^ 2at^
are t-^y =^x-\- at^,
and t.^y — x-\- at^.
The point of -intersection of these two tangents is clearly
{<X^1^2} ^(^1 + ^2)}-
The point whose coordinates are (a<^, 2a£) may, for
brevity, be called the point " tT
In the following articles we shall prove some important
properties of the parabola making use of the above substi-
tution.
200 COORDINATE GEOMETRY.
230. If the tangents at P and Q meet in T, prove that
(1) TP and TQ subtend equal angles at the focus Sy
(2) ST^=SP.SQ,
and (3) the triangles SPT and STQ are similar.
Let P be the point {at^^, 2at^), and Q be the
point (at^^, 2at2), so that (Art. 229) T is the point
{atjt^, a(ii + t2)}-
2at
(1) The equation to SP is y= ^ ^ {x - a),
i. e. [t-^ -l)y-2t-^x + 2at-^ = 0.
The perpendicular, TU, from T on this
straight line
a{t ^^- 1) (fi + 1^) - 2^1 . ati tg + 2at^ ^ {t^^ - t^^t^) + (^i - ^
^^(fi-io)-
Similarly TC/' has the same numerical value.
The angles PST and QST are therefore equal.
(2) By Art. 202 we have SP=a{l + 1^^) and SQ = a(l + f.^).
Also ST^ = (af 1^2 -a)^ + a^{t^ + t^y-
= a^[tj;'t^^ + t^^ + t^^+l-\ = a^l + tj^){l + t^-^).
Hence ST^ = SP.SQ.
ST SO
(3) Since — ^ = ^, and the angles TSP and T^SQ are equal, the
oP ol
triangles SPT and STQ are similar, so that
Z /SQT^ z ^TP and Z ^rQ= z 5fPT.
231. 27ze area of the triangle formed by three points on a
parabola is twice the area of the triangle formed by the tangents at
these points.
Let the three points on the parabola be
{at-^, 2at-^, [at^, 2at^, and {at^^, 2at^.
The area of the triangle formed by these points, by Art. 25,
= i [at-^ {2af2 - Saig) + at^ {2at^ - 2at-^) + at^ {2at-^ - 2at^)'\
The intersections of the tangents at these points are (Art. 229)
the points
{at^t.^, a[t^ + t^\, {at^t^, a{t^-\-tT^], and [at^t^, a{t^-\-t^].
The area of the triangle formed by these three points
= \ {at^t^ {at^ - at^) + atj:-^ [at-^ - at^ + at-^t^ [at^-at^]
=W{t^-ts){t^-t^){t^-t^).
The first of these areas is double the second.
MISCELLANEOUS EXAMPLES. 201
232. The circle circumscribing the triangle formed by any three
tangents to a parabola passes through the focus.
Let P, Q, and R be the points at which the tangents are drawn
and let their coordinates be
{atj^, 2afj), (aig^ ^at^), and {at^^, 2at^).
As in Art. 229, the tangents at Q and R intersect in the point
{at^t^, a{t2 + t^)}.
Similarly, the other pairs of tangents meet at the points
{afgfi, ^(ig + ij)} and {at^t.^, a{tj^ + t2)}.
Let the equation to the circle be
x^ + y^ + 2gx + 2fy + c = (1).
Since it passes through the above three points, we have
a\%^ + a2 (<2 + t.^)^ + 2gat^t^ + 2/a (f g + fg) + c = (2),
aH^t^ + a2 {t^ + 1^'2 4- 2gat^t-^ + 2/a (tg + f^) + c = (3),
and aH-^t^ + a^ («i + ^ ^ + 2gat-^t^ + 2fa{t^ + t^ + c = Q (4).
Subtracting (3) from (2) and dividing by a {t^ - t^), we have
a {fgS (fi + fa) + <i + ^3 + 2*3} + 2gt^ + 2/= 0.
Similarly, from (3) and (4), we have
«{«i'(*2 + «3) + «2 + *3 + 2«i} + 25rfi + 2/=0.
From these two equations we have
2g=-a{l-\- t^ts + t^t-i + tjt^) and 2f== -alt-^ + t.^ + t^- t^t^ts].
Substituting these values in (2), we obtain
c = a'^{t^ts + tstj^ + t-it2).
The equation to the circle is therefore
x^ + y^- ax (1 + £2*3 + *3^i + ^1^2) ~ ^^y ih + ^2 + ^3 ~ *i^2^3)
+ a- (t^t^ + t^t^ + tj^t^) = 0,
which clearly goes through the focus {a, 0).
233. If be any point on the axis and POP' be any chord
passing through 0, and if PM and P'M' be the ordinates of P and P',
prove that AM . AM' = AO\ and PM . P'M'= - 4a . AO.
Let O be the point {h, 0), and let P and P' be the points
(afj2, 2atj) and {at^, 2at^.
The equation to PP' is, by Art. 229,
{t^ + t-^y -2x = 2at^t^.
If this pass through the point [h, 0), we have
-2h = 2at-^t^,
202 COORDINATE GEOMETRY.
Hence AM. A M' =aU^. aU^ =a^.- = K^=A 0^-
and PM . PM' = 2a\ . 2atc^ . =4^2 (--)=- 4a . AO.
icus, AO = a, and
Cor. If be the focus, ^0 = a, and we have
1
The points {at^, 2at-^ and ( — ^ , ) are therefore at the ends
of a focal chord.
234. To prove that the orthocentre of any triangle formed by
three tangents to a parabola lies on the directrix.
Let the equations to the three tangents be
y = nhx+— (1),
y=m^ + -- (2),
nt
and y = mgX-\ (3).
m.
The point of intersection of (2) and (3) is found, by solving them,
to be
The equation to the straight line through this point perpendicular
to (1) is (Art. 69)
y-a{— + —) = \ X ~ ,
X rl 1 a ~\
I.e. ?/+— =a — + — + (4).
Similarly, the equation to the straight line through the intersection
of (3) and (1) perpendicular to (2) is
X ( \ 1 d \
y + — =a _ + _+ , 5)
and the equation to the straight line through the intersection of (1)
and (2) perpendicular to (3) is
X f \ \ ci \
?/ + — = a — +— + 6.
wig \^\ ^2 ni-^m^m^J
The point which is common to the straight lines (4), (5), and (6),
EXAMPLES. ONE VARIABLE. 203
i.e. the orthocentre of the triangle, is easily seen to be the point
whose coordinates are
/111 1
x=-a^ y = a[—'-\ ! +
and this point lies on the directrix.
EXAMPLES. XXVIII.
1. If w be the angle which a focal chord of a parabola makes with
the axis, prove that the length of the chord is 4a cosec'-^ w and that the
perpendicular on it from the vertex is a sin w.
2. A point on a parabola, the foot of the perpendicular from it
upon the directrix, and the focus are the vertices of an equilateral
triangle. Prove that the focal distance of the point is equal to the
latus rectum.
3. Prove that the semi-latus-rectum is a harmonic mean between
the segments of any focal chord.
4. If T be any point on the tangent at any point P of a parabola,
and if TL be perpendicular to the focal radius SP and TN be perpen-
dicular to the directrix, prove that SL = TN.
Hence obtain a geometrical construction for the pair of tangents
drawn to the parabola from any point T.
5. Prove that on the axis of any parabola there is a certain point
K which has the property that, if a chord PQ of the parabola be drawn
through it, then
1 1
PK^'^'QK^
is the same for all positions of the chord.
6. The normal at the point (at-^, 2atj) meets the parabola again
in the point {aU^, 2at2) ; prove that
2
H
7. A chord is a normal to a parabola and is inclined at an angle
d to the axis ; prove that the area of the triangle formed by it and
the tangents at its extremities is 4a'^ sec^ 6 cosec^ 0.
8. If PQ be a normal chord of the parabola and if S be the focus,
prove that the locus of the centroid of the triangle SPQ is the curve
36a2/2 (3a; - 5a) - 81^^= 128a'^.
9. Prove that the length of the intercept on the normal at the
point {at^, 2at) made by the circle which is described on the focal
distance of the given point as diameter is a f^/l + 1^.
204 COORDINATE GEOMETRY. [EXS.
10. Prove that the area of the triangle formed by the normals to
the parabola at the points {at^, Satj), {at^, 2at^ and [at^y Satg) is
\ («2 - h) («3 - h) ih - h) ih + *2 + «3)'.
11. Prove that the normal chord at the point whose ordinate
is equal to its abscissa subtends a right angle at the focus.
12. A chord of a parabola passes through a point on the axis
(outside the parabola) whose distance from the vertex is half the
latus rectum ; prove that the normals at its extremities meet on the
curve.
13. The normal at a point P of a parabola meets the curve
again in Q, and T is the pole of PQ; shew that T lies on the diameter
passing through the other end of the focal chord passing through P,
and that PT is bisected by the directrix.
14. If from the vertex of a parabola a pair of chords be drawn at
right angles to one another and with these chords as adjacent sides a
rectangle be made, prove that the locus of the further angle of the
rectangle is the parabola
?/2 = 4a (a; -8a).
15. A series of chords is drawn so that their projections on a
straight line which is inclined at an angle a to the axis are all of
constant length c ; prove that the locus of their middle point is the
curve
{y^-iax) {y qoq a -\-2a Bin of + a^c^ = 0.
16. Prove that the locus of the poles of chords which subtend a
right angle at a fixed point (/i, /c) is
ax^ - %2 + (4a'^ + 2ali) x - 2ahj + a {h^ + A;-) = 0.
17. Prove that the locus of the middle points of all tangents
drawn from points on the directrix to the parabola is
y^{2x + a) = a{dx + a)-.
18. Prove that the orthocentres of the triangles formed by three
tangents and the corresponding three normals to a parabola are
equidistant from the axis.
19. T is the pole of the chord PQ ; prove that the perpendiculars
from P, T, and Q upon any tangent to the parabola are in geometrical
progression.
20. If '^1 and r^ be the lengths of radii vectores of the parabola
which are drawn at right angles to one another from the vertex, prove
that
rji^r2^=16a2{ri^ + r2^).
21. A parabola touches the sides of a triangle ABC in the points
D, E, and F respectively ; if DE and DF cut the diameter through the
point A inb and c respectively, prove that Bb and Cc are parallel.
XXVIII.] EXAMPLES. ONE VARIABLE. 205
22. Prove that all circles described on focal radii as diameters
touch the directrix of the curve, and that all circles on focal radii as
diameters touch the tangent at the vertex,
23. -A- circle is described on a focal chord as diameter ; if m be the
tangent of the inclination of the chord to the axis, prove that the
equation to the circle is
\ m^J m
24. LOL' and il/Oilf'are two chords of a parabola passing through
a point on its axis. Prove that the radical axis of the circles
described on LL' and MM' as diameters passes through the vertex of
the parabola.
25. -A- circle and a parabola intersect in four points; shew that the
algebraic sum of the ordinates of the four points is zero.
Shew also that the line joining one pair of these four points and
the line joining the other pair are equally inclined to the axis.
26. Circles are drawn through the vertex of the parabola to cut
the parabola orthogonally at the other point of intersection. Prove
that the locus of the centres of the circles is the curve
2i/2 (22/2 + a;2 - Viax) = ax {%x - 4a)2.
27. Prove that the equation to the circle passing through the
points {at^, 2,at^ and (2a<2^, 2at^ and the intersection of the tan-
gents to the parabola at these points is
ic2 + 2/2 - ax [(«! + «2)^ + 2] - aij [t^ + 1^ (1 - t-^ t^) + a2 1^ t^ (2 - 1-^ t^ = 0.
28. TP and TQ are tangents to the parabola and the normals at P
and Q meet at a point R on the curve ; prove that the centre of the
circle circumscribing the triangle TPQ Kes on the parabola
2y^ = a{x — a).
29. Through the vertex A of the parabola ?/2 = 4aa; two chords AP
and AQ are drawn, and the circles on AP and ^Q as diameters
intersect in R. Prove that, if 6-^, 6^,^ and be the angles made with
the axis by the tangents at P and Q and by AR, then
cot ^i + cot ^2 + 2 tan0 = O.
30. A- parabola is drawn such that each vertex of a given triangle
is the pole of the opposite side ; shew that the focus of the parabola
lies on the nine-point circle of the triangle, and that the orthocentre of
the triangle formed by joining the middle points of the sides lies on
the directrix.
CHAPTER XL
THE PARABOLA {continued).
[On a first reading of this Chapter, the student may, with
advantage, omit from Art. 239 to the end.]
Some examples of Loci connected with the
Parabola.
235. Ex. 1. Find the locus of the intersection of tangents to the
•parabola y^ = 4iax, the angle hetioeen them being always a given angle a.
The straight line y ^mx -] — is always a tangent to the parabola.
If it pass through the point T {h, h) we
have
m^h-mk + a = (1).
If mj and Wg be the roots of this equation -»-
we have (by Art, 2) (/ik)
h
and
a
.(2),
•(3),
and the equations to TP and TQ are then
a
and y — nux -^ — - .
1 m^
y—miX + — »^« y—ni,^.
Hence, by Art. 66, we have
m^ - m^ _ ijjm-^ + m^)^ - ^m^ m^
tana
1 + m^m^
1 + WljWg
/
k^ _4a
a + h
h
, by (2) and (3).
THE PARABOLA. LOCI. 207
.-. k^-4:ah = {a + h)^t&n^a.
Hence the coordinates of the point T always satisfy the equation
7/2 - 4kax = {a + x)^ tan^ a.
We shall find in a later chapter that this curve is a hyperbola.
As a particular case let the tangents intersect at right angles, so
that m^^= - 1.
From (3) we then have h= -a, so that in this case the point T lies
on the straight line x= -a, which is the directrix.
Hence the locus of the point of intersection of tangents, which cut
at right angles, is the directrix.
Ex. 2. Prove that the locus of the poles of chords which are normal
to the parabola y^ = 4:ax is the curve
2/2(a; + 2a) + 4a3 = 0.
Let PQ be a chord which is normal at P. Its equation is then
y = mx-2am-am^ (1).
Let the tangents at P and Q. intersect in T, whose coordinates are
h and k, so that we require the locus of T.
Since PQ is the polar of the point {h, k) its equation is
yk=2a{x + h) (2).
Now the equations (1) and (2) represent the same straight line, so
that they must be equivalent. Hence
m = — , and - 2am - am^ = —j— .
Eliminating m, i. e. substituting the value of m from the first of
these equations in the second, we have
4a2 8a^ _ 2ah
~T~'W~~k''
i.e. ^ k^{h + 2a} + 4a^=0,
The locus of the point T is therefore
2/2(a; + 2a) + 4a3=0.
Ex. 3. Find the locus of the middle points of chords of a parabola
which subtend a right angle at the vertex, and prove that these chords all
pass through a fixed point on the axis of the curve.
208
COORDINATE GEOMETRY.
First Method. Let PQ be any sucli chord, and let its equation be
y = mx + c (1).
The lines joining the vertex with the
points of intersection of this straight line y
with the parabola
y^=^ax (2),
are given by the equation A
y^c = 4ax {y - mx). (Art. 122)
These straight lines are at right angles if
c + 4am=:0. (Art. Ill)
Substituting this value of c in (1), the
equation to FQ is
y = m {x - 4a)
This straight line cuts the axis of a; at a constant distance 4a from
the vertex, i.e. AA' = 4a.
If the middle point oi PQ be {h, k) we have, by Art. 220,
2a
(3).
li=:
(4).
.(5).
m
Also the point {h, k) lies on (3), so that we have
k = m{h-4a)
If between (4) and (5) we eliminate m, we have
kJ-^{h-4a),
i.e. k^=1a{h-4ia),
so that {h, k) always lies on the parabola
2/2 = 2a (a; -4a).
This is a parabola one half the size of the original, and whose
vertex is at the point A' through which all the chords pass.
Second Method. Let P be the point (afj^, 2at^ and Q be the point
{at^, 2aio).
The tangents of the inclinations of AP and AQ io the axis are
— and — .
1 2
Since AP and AQ are at right angles, therefore
*1 *2
.(6).
i.e. tih= -4
As in Art. 229 the equation to PQ is
{ty^ + t^y=:2x + 2at^t^ (7).
THE PARABOLA. LOCI. 209
This meets the axis of ic at a distance -atit2,i.e., by (6), 4a, from
the origin.
Also, {h, k) being the middle point of PQ, we have
and 2k = 2a{tj^ + t2).
Hence k- - 2ah = a^ (f^ + 1^)^ - a^ {tj^ + 1^^)
= 2a\t2= -8a2,
so that the locus of (h, k) is, as before, the parabola
y^ — 2a{x- 4a).
Tbird Method. The equation to the chord which is bisected at
the point {h, k) is, by Art. 221,
k{y-k) = 2a{x-h),
i. e. ky - 2ax=k^ - 2ah (8).
As in Art. 122 the equation to the straight lines joining its points
of intersection with the parabola to the vertex is
{k^ - 2ah) 2/2 = iax {ky - 2ax).
These lines are at right angles if
{k^-2ah) + 8a^ = 0.
Hence the locus as before.
Also the equation (8) becomes
ky - 2ax = - 8a^.
This straight line always goes through the point (4a, 0).
EXAMPLES. XXIX.
From an external point P tangents are drawn to the parabola ; find
the equation to the locus of P when these tangents make angles 6^ and
$2 with the axis, such that
1. tan dj^ + tan 6^ is constant ( = 6).
2. tan 6^ tan d^ is constant { = c).
3. cot 6^ + cot $2 is constant { = d).
4. di + ^2 is constant ( = 2a).
5 . tan^ 6-^ + tan^ ^g is constant ( = X) .
6. cos ^1 cos $2 is constant ( = /u),
L. 14
210 COORDINATE GEOMETRY. [ExS.
7. Two tangents to a parabola meet at an angle of 45° ; prove that
the locus of their point of intersection is the curve
y^ - \.ax = {x + of.
If they meet at an angle of 60°, prove that the locus is
2/2-3a;2-10aic-3a2^0.
8. A pair of tangents are drawn which are equally inclined to a
straight line whose inclination to the axis is a ; prove that the locus
of their point of intersection is the straight line
y = {x-a) tan 2a.
9. Prove that the locus of the point of intersection of two tangents
which intercept a given distance 4c on the tangent at the vertex is an
equal parabola.
10. Shew that the locus of the point of intersection of two tangents,
which with the tangent at the vertex form a triangle of constant area
c^, is the curve x^ [y^ - 4:ax)=4:C^a^.
11. If the normals at P and Q meet on the parabola, prove that
the point of intersection of the tangents at P and Q lies either on a
certain straight line, which is parallel to the tangent at the vertex, or
on the curve whose equation is y^ {x + 2a) + 4a^ = 0.
12. Two tangents to a parabola intercept on a fixed tangent
segments whose product is constant ; prove that the locus of their
point of intersection is a straight line.
13. Shew that the locus of the poles of chords which subtend a
constant angle a at the vertex is the curve
(x + A.af=4: cot^ a (t/^ - 4aa;).
14. In the preceding question if the constant angle be a right angle
the locus is a straight line perpendicular to the axis.
15. A point P is such that the straight line drawn through it
perpendicular to its polar with respect to the parabola y^=4:ax touches
the parabola x'^ = Aby. Prove that its locus is the straight line
2ax + by + 4:a^=0.
16. Two equal parabolas, A and B, have the same vertex and axis
but have their concavities turned in opposite directions ; prove that
the locus of poles with respect to B of tangents to A is the parabola A.
17. Prove that the locus of the poles of tangents to the parabola
y^—Aax with respect to the circle x^ + y'^=2ax is the circle x^ + y^=ax.
18. Shew the locus of the poles of tangents to the parabola
y^—Aax with respect to the parabola y^=4bx is the parabola
y'^= — X.
XXIX.] THREE NORMALS FROM ANY POINT.
211
Find the locus of the middle points of chords of the parabola
which
19. pass through the focus.
20. pass through the fixed point (/?., k).
21. are normal to the curve.
22. subtend a constant angle a at the vertex.
23. are of given length I.
24. are such that the normals at their extremities meet on the
parabola.
25. Through each point of the straight line x = niy + h is drawn
the chord of the parabola y^=4iax which is bisected at the point;
prove that it always touches the parahola
{y - 2am)^=8d {x - h).
26. Two parabolas have the same axis and tangents are drawn to
the second from points on the first ; prove that the locus of the middle
points of the chords of contact with the second parabola all lie on a
fixed parabola.
27. Prove that the locus of the feet of the perpendiculars drawn
from the vertex of the parabola upon chords, which subtend an angle
of 45° at the vertex, is the curve
r2 - 24ar cos 6 + 16a^ cos 29 = 0.
236. To prove that, in general, three normals can be
drawn from any point to the parahola and that the algebraic
sum, of the ordinates of the feet of these three normals is
zero.
The straight line
y — mx — 2am — amP
is, by Art. 208, a normal to the
parabola at the points whose coordi-
nates are
arn^ and — 2am (2).
If this normal passes through
the fixed point 0, vi^hose coordinates
are h and k, we have
k = mh — 2am, — am,\
%.e.
arn? + (2a — h) m + k
(1)
212 COORDINATE GEOMETRY.
This equation, being of the third degree, has three
roots, real or imaginary. Corresponding to each of these
roots, we have, on substitution in (1), the equation to a
normal which passes through the point 0.
Hence three normals, real or imaginary, pass through
any point 0.
If m^, iiu, and m^ be the roots of the equation (3), we
have
m^ + m.^ + m^ = 0.
If the ordinates of the feet of these normals be 2/^, y^,
and 2/3, we then have, by (2),
2/1 + 2/2 + 2/3 = - 2o^ {m^ + ^2 + m^) = 0.
Hence the second part of the proposition.
We shall find, in a subsequent chapter, that, for certain
positions of the point 0, all three normals are real ; for
other positions of 0, one normal only will be real, and the
other two imaginary.
237. Ex. Find the locus of a point which is such that (a) two of
the normals drawn from it to the parabola are at right angles,
(j8) the three normals through it cut the axis in points whose distances
from, the vertex are in arithjnetical progression.
Any normal is y=mx-2am-am^, and this passes through the
point {h, k), if
am^ + {2a-h)m+k = (1).
If then nil , m^, and m^ be the roots, we have, by Art. 2,
m-y + m^ + m^ = , (2) ,
m^m^ + m^mi + mim^= , (3),
k
and mim2m^= — (4).
(a) If two of the normals, say m^ and m^^ , be at right angles, we
k
have wiim2=-l, and hence, from (4), m^—-.
k
The quantity - is therefore a root of (1) and hence, by substitution,
Of
we have
-+{2a-h)~ + k = 0,
a^ ^ a
i.e. k^=a{h-3a).
THREE NORMALS. EXAMPLES. 213
The locus of the point (/^, k) is therefore the parabola y'^ = a{x- 3a)
whose vertex is the point (3a, 0) and whose latus rectum is one-quarter
that of the given parabola.
The student should draw the figure of both parabolas.
(/3) The normal y = mx - 2am - am^ meets the axis of a; at a point
whose distance from the vertex is 2a + aw^. The conditions of the
question then give
(2a + awij^) + (2a + am^^) = 2 (2a + am^^) ,
i.e. m-^ + m.^ = 2m<^ (5).
If we eliminate m^, m^, and m^ from the equations (2), (3), (4),
and (5) we shall have a relation between h and k.
From (2) and (3), we have
= vi^m.^ + m^ (wii + m^) = m^m^ - m^^ (6).
Also, (5) and (2) give
2m2^ = {v\ + wig)^ - Im^m.^ — m.^ - Im-^^ ,
i.e. m^ + 2m^m^ =0 (7).
Solving (6) and (7), we have
2a - 7t , „ ^ 2a-h
vi.mo= —z — , and ot^-^ - 2 x —r — .
^ •* 3a -^ 3a
Substituting these values in (4), we have
2 a- fe / 2a-h_ k
3a V ~Sa^ ~ ~ a '
i.e. 27afc2 = 2(7i-2a)3,
so that the required locus is
27ai/2=2(a;-2a)3.
238. Ex. If the normals at three points P, Q, and R meet in a
point and S be the focus, prove that SP . SQ . SR = a . S0\
As in the previous question we know that the normals at the
points (ayn-^^, -2amj), {ain.^,-2am2^ and {am.^,-2am^) meet in the
point (A, k) if
mj + ^2 + m3 = (1),
2a ~ h
m^vi.^ + m^m-^ + m^m^ — — (2) ,
and mnW.,??io = — (3).
'■ - " a ^ '
By Art. 202 we have
SP=a{l+m^), SQ = a{l + m^-), and SR = a{l + m^^y
214 COORDINATE GEOMETRY.
Hence ^P-^Q-^^ ^ ^1 _^ ^^^2) (i + ^^^2) (i + ,,,^.2)
= 1 + (Wj^ + m^^ + m^) + [m^m^ + m^m^ + m^m^) + m^m^m.^.
Also, from (1) and (2),' we have
m^ + 7n2^ + Wg^ = (wi + 7712 + wis)^ - 2 (?»oW3 + ma?^! + m^mo)
_ h-2a
a
and
= (^y,by (l)and(2).
„ SP.SQ.SR ^ li-2a fli-2ay ¥■
Hence ^ = 1 + 2 + )+-
i,e. SP.SQ.SR = SO^.a.
EXAMPLES. XXX.
Find the locus of a point O when the three normals drawn from
it are such that
1. two of them make complementary angles with the axis.
2. two of them make angles with the axis the product of whose
tangents is 2.
3. one bisects the angle between the other two.
4. two of them make equal angles with the given line y = vix + c.
5. the sum of the three angles made by them with the axis is
constant.
6. the area of the triangle formed by their feet is constant.
7. the line joining the feet of two of them is always in a given
direction.
The normals at three points P, Q, and R of the parabola y^ = 4:ax
meet in a point whose coordinates are h and k ; prove that
8. the centroid of the triangle PQR lies on the axis.
9. the point and the orthocentre of the triangle formed by the
tangents at P, Q, and R are equidistant from the axis.
[EXS. XXX.] THREE NORMALS. EXAMPLES. 215
10. if OP and OQ make complementary angles with the axis, then
the tangent at R is parallel to SO.
11. the sum of the intercepts which the normals cut off from the
axis is 2{h + a).
12. the sum of the squares of the sides of the triangle PQR is
equal to 2{h-2a){h + 10a).
13. the circle circumscribing the triangle PQR goes through the
vertex and its equation is 2x^ + 2y- -2x{h + 2a) -ky = 0.
14. if P be fixed, then QR is fixed in direction and the locus of
the centre of the circle circumscribing PQR is a straight line.
15. Three normals are drawn to the parabola y^ = 4:ax cos a from
any point lying on the straight line y = h sin a. Prove that the locus
of the orthocentre of the triangles formed by the corresponding tan-
gents is the curve -2 + t^ ~ ■'■' ^^^ angle a being variable.
16. Prove that the sum of the angles which the three normals,
drawn from any point 0, make with the axis exceeds the angle which
the focal distance of O makes with the axis by a multiple of tt.
17. Two of the normals drawn from a point to the curve make
complementary angles with the axis ; prove that the locus of and
the curve which is touched by its polar are parabolas such that their
latera recta and that of the original parabola form a geometrical
progression. Sketch the three curves.
18. Prove that the normals at the points, where the straight line
lx + my = l meets the parabola, meet on the normal at the point
/4am2 4am \ „ , , . .
( , — r— j of the parabola.
19. If the normals at the three points P, Q, and R meet in a point
and if PP', QQ', and RR' be chords paraUel to QR, RP, and PQ
respectively, prove that the normals at P\ Q', and R' also meet in a
point.
20. If the normals drawn from any point to the parabola cut the
line x=2a in points whose ordinates are in arithmetical progres-
sion, prove that the tangents of the angles which the normals make
with the axis are in geometrical progression.
21. PG, the normal at P to a parabola, cuts the axis in G and is
produced to Q so that GQ = ^PG \ prove that the other normals
which pass through Q intersect at right angles.
22. Prove that the equation to the circle, which passes through the
focus and touches the parabola y^ = 4.ax at the point [at^, 2at), is
x^- + y^- ax {St^ + 1) - ay {St - t^) + SaH^=0.
Prove also that the locus of its centre is the curve
27 ay- = {2x - a) {x - 5a)-.
216 COORDINATE GEOMETRY. [Exs. XXX.]
23. Shew that three circles can be drawn to touch a parabola and
also to touch at the focus a given straight line passing through the
focus, and prove that the tangents at the point of contact with the
parabola form an equilateral triangle.
24. Through a point P are drawn tangents FQ and PR to a
parabola and circles are drawn through the focus to touch the para-
bola in Q and iJ respectively ; prove that the common chord of these
circles passes through the centroid of the triangle FQR.
25. Prove that the locus of the centre of the circle, which passes
through the vertex of a parabola and through its intersections with a
normal chord, is the parabola 2y'^=iax-a^.
26. -A- circle is described whose centre is the vertex and whose
diameter is three-quarters of the latus rectum of a parabola ; prove
that the common chord of the circle and parabola bisects the distance
between the vertex and the focus.
27. Prove that the sum of the angles which the four common
tangents to a parabola and a circle make with the axis is equal to
mr + 2a, where a is the angle which the radius from the focus to the
centre of the circle makes with the axis and n is an integer.
28. -P^ and QR are chords of a parabola which are normals at P
and Q. Prove that two of the common chords of the parabola and
the circle circumscribing the triangle PRQ meet on the directrix.
29. The two parabolas y^ = 4a(x-l) and x^ = 'ia(y-l') always
touch one another, the quantities I and V being both variable ; prove
that the locus of their point of contact is the curve xy='ia^.
30. A parabola, of latus rectum i^, moves so as always to touch an
equal parabola, their axes being parallel ; prove that the locus of their
point of contact is another parabola whose latus rectum is 21.
31. The sides of a triangle touch a parabola, and two of its angular
points lie on another parabola with its axis in the same direction ;
prove that the locus of the third angular point is another parabola.
239. In Art. 197 we obtained the simplest possible
form of the equation to a parabola.
We shall now transform the origin and axes in the
most general manner.
Let the new origin have as coordinates [h, k), and let
the new axis of x be inclined at to the original axis, and
let the new angle between the axes be o>'.
PAEABOLA. TWO TANGENTS AS AXES. 217
By Art. 133 we have for x and y to substitute
X cos B -^y cos (w' + ^) + A,
and X sin Q + y sin (to' + ^) + ^
respectively.
The equation of Art. 197 then becomes
{x sin B-\-y sin (o>' + ^) + ^|^ = 4a {x cos ^ + y cos (w' + 6^) + A},
i.e.
\x sin ^ + 2/ sin (<o' + ^)}^ + 2£c {A; sin ^ - 2a cos ^}
+ 1y {h sin (w' + ^) - 2a cos (w' + ^)} + ^^^ _ 4^;^ ^ q
■••••••(l)-
This equation is therefore the most general form of the
equation to a parabola.
We notice that in it the terms of the second degree
always form a perfect square.
240. To find the equation to a 'parabola, any two
tangents to it being the axes of coordinates and the points of
contact being distant a and b from the origin.
By the last article the most general form of the equa-
tion to any parabola is
{Ax-^Byf-v2gx+2fy-^c = (1).
This meets the axis of x in points whose abscissae are
given by
^ V + ^gx + c = (2).
If the parabola touch the axis of x at a distance a from
the origin, this equation must be equivalent to
A^{x-af = (3).
Comparing equations (2) and (3), we have
g- — A^a, and c = A^a^ (4).
Similarly, since the parabola is to touch the axis of y
at a distance b from the origin, we have
f=-B%, and c=^B''b' (5).
218 COORDINATE GEOMETRY.
From (4) and (5), equating the values of c, we have
sothat B = ±A^ (6).
Taking the negative sign, we have
B = -A'^, g = -'A\ f^-A'j, and G=A'a\
Substituting these values in (1) we have, as the required
equation,
t.e.
('?_iy_?^_^+i=o (7).
\a 0/ a
This equation can be written in the form
\a 0/ \a 0/ ao
ah'
yivi=i <«)•
a
I.e. I ^ / -+ /v/ ^j =lj
^.e.
[The radical signs in (8) can clearly have both the positive and
negative signs prefixed. The different equations thus obtained corre-
spond to different portions of the curve. In the figure of Art. 243,
the abscissa of any point on the portion PAQ is <a, and the ordinate
< h, so that for this portion of the curve we must take both signs
positive. For the part beyond P the abscissa is > a, and - > r , so
that the signs must be + and - . For the part beyond Q the
fit n^
ordinate is >&, and v>-, so that the signs must be - and +.
a
There is clearly no part of the cui-ve corresponding to two negative
signs.]
PARABOLA. TWO TANGENTS AS AXES. 219
241. If in the previous article we took the positive
sign in (6), the equation would reduce to
(- + I-) _2--^ + l = 0,
\a J a
This gives us (Fig., Art. 243) the pair of coincident
straight lines PQ. This pair of coincident straight lines is
also a conic meeting the axes in two coincident points at P
and Q, but is not the parabola required.
242. To find the equation to the tangent at any point
(x', y') of the parabola
Let (cc", 2/") be any point on the curve close to {x ^ y).
The equation to the line joining these two points is
y - 2/' = |iJ~' («-«') (1)-
But, since these points lie on the curve, we have
Ix
sl ~a
^ . /y 1 M' , Ivi
+ ^/| = l = Vl^-^/y (2).
so that ^gz^ = -^ (3).
sjx' — s]x s]a
The equation (1) is therefore
/ _ sly" - -Jy sly" + ^ y , _ ,.
y y ~ I— i~, j-r, , r,\^ ^n
sJX — \JX \JX + sjX !
or, by (3),
; , sjh sly" + sjy , ,. ...
y-y=--,- -h, — F=X^-^) (4).
sja six + ycc
220 COORDINATE GEOMETRY.
The equation to the tangent at {x\ y') is then obtained
by putting x" = x and y" — y', and is
Jh sly'
y-y =--7- -7^ (x-x),
s/a \Jx
%.e.
slax''^ slhy'~ \' a ^ \ h~^ ^^^*
This is the required equation.
[In the foregoing we have assumed that [x', y') Hes on the portion
PAQ (Fig., Art. 243). If it lie on either of the other portions the
proper signs must be affixed to the radicals, as in Art. 240.]
CO u
Ex. To find the condition that the straight line ^+ - = lmay be a
tangent.
This line will be the same as (5), if
f = Jax' and g=Jhy',
1^' f fv' Q
SO that s. —~-i and m.I^ — t-
'V '' « \' h
Hence - + ^ = 1.
a
This is the required condition; also, since x'=— and ^'=— ,
the point of contact of the given line is ( "— , ^ J .
Similarly, the straight line lx + my = n will touch the parabola if
n ** _i
al bm '
243. To find the focus of the j^arabola
a W h
Let ^S' be the focus, the origin, and P and Q the
points of contact of the parabola with the axes.
Since, by Art. 230, the triangles OSP and QSO are
similar, the angle SOP- angle SQO.
Hence if we describe a circle through 0, Q, and S, then,
by Euc. III. 32, OP is the tangent to it at P.
PARABOLA. TWO TANGENTS AS AXES. 221
Hence S lies on the circle passing through the origin
0, the point Q, (0, b), and touching the axis of x at the
origin.
P X
The equation to this circle is
x^ + 2xy cos oi +y^ = by ( 1 ).
Similarly, since z SOQ — L SPO^ S will lie on the circle
through and P and touching the axis of y at the origin,
i.e. on the circle
x^ + 2xy cos (0 +y'^ — ax (2).
The intersections of (1) and (2) give the point required.
On solving (1) and (2), we have as the focus the point
air" a^h
qP' + 2ah cos <o + 6" ' ffl^ + 2ah cos <o + 6'
244. To find the equation to the axis.
If V be the middle point of PQ^ we know, by Art. 223,
that F is parallel to the axis.
Now V is the point ( - , - ) .
Hence the equation to F is y = -x.
The equation to the axis (a line through S parallel to
F) is therefore
a^h h ( ah^ \
If r= — I £C — ■ I
a^ + 2ab cos w +b^ a\ a^ + 2ab cos w + 6v '
, ab (a? — b^)
*• ^- ay-bx= ~ — —\ ^ — - .
a^ + lab cos o) + 6"
222 COORDINATE GEOMETRY.
245 . To find the equation to the directrix.
If we find the point of intersection of OP and a
tangent perpendicular to OP, this point will (Art. 211, y)
be on the directrix.
Similarly we can obtain the point on OQ which is on
the directrix.
A straight line through the point (/, 0) perpendicular
to OX is
y = in{x —f), where (Art. 93) 1 +7n cos o> = 0.
The equation to this perpendicular straight line is
then
x+ 2/ cos (0 =y (1).
This straight line touches the parabola if (Art. 242)
/ / 1 • -I. ^' «^ cos
(X)
a b cos o) ' ' a+ b cos to
ab cos (0
The point ( — — = , ) therefore lies on the directrix.
\a + b cos <o /
^. ., , 1 . //^ ab cos, (0 \ .
Similarly the point ( 0, , ) is on it.
•^ ^ \ b +a cos (0/
The equation to the directrix is therefore
x{a + b cos 0)) +1/ (b + a cos w) = ab cos oo (2).
The latus rectum being twice the perpendicular distance
of the focus from the directrix — twice the distance of the
point
/ ab^ a'^b \
\aP + 2ab cos oo + 5^ ' a"^ + 2ab cos w + b^J
from the straight line (2)
4<x^6^ sin^ w
~ (a^ + 2ab cos oo + 6^)t '
by Art. 96, after some reduction.
246. To find the coordinates of the vertex aiid the
equation to the tangent at the vertex.
PARABOLA. TWO TANGENTS AS AXES.
223
The vertex is the intersection of the axis and the curve,
i.e. its coordinates are given by
H X a? — h^ .
and by
i.e. by
yy
h
a? + 2ah cos <o + 6^
X
fx
\a
-^-1+1 =
a
From (1) a,nd (2), we have
x =
1
O 7 <>
a" — 0-
a?' + lab cos a> + 6^
alP" {h ■\- a cos w)^
.(Art. 240),
(2).
{ci? + 2a& cos (0 + IP'Y '
Similarly y
a^b (a + b cos tof
(a? + 2ab cos w + b^Y '
These are the coordinates of the vertex.
The tangent at the vertex being parallel to the directrix,
its equation is
«6^ {b + a cos uif
(a + b cos co)
X —
a^ + 2ab cos oo + b^f
,-, . r a%(a + b cos o>)"^ H .
+ (6 + « cos co) 2/ - j—^ )r-j~ -^ r = 0,
^ L (« + 2ao cos (o + ¥fj
^.e.
— +
2/
«.6
b + a cos (0 a + 5 cos co a^ + 2ab cos co + 6^
EXAMPLES. XXXI.
1. If a parabola, whose latus rectum is 4c, slide between two
rectangular axes, prove that the locus of its focus is x'^y^=c^ {x^ + y^),
and that the curve traced out by its vertex is
2. Parabolas are drawn to touch two given rectangular axes and
their foci are all at a constant distance c from the origin. Prove that
the locus of the vertices of these parabolas is the curve
x^ + y^=c'"
224 COOKDINATE GEOMETRY. [ExS. XXXI.]
3. The axes being rectangular, prove that the locus of the focus
of the parabola (- + r-l) = — t j ^ 3,nd h being variables such
\a j ah
that db = c^, is the curve {x'^-\-']ff — e^xy.
4. Parabolas are drawn to touch two given straight lines which
are inclined at an angle w ; if the chords of contact all pass through
a fixed point, prove that
(1) their directrices all pass through another fixed point, and
(2) their foci all lie on a circle which goes through the intersection of
the two given straight lines.
5. A parabola touches two given straight lines at given points ;
prove that the locus of the middle point of the portion of any tangent
which is intercepted between the given straight lines is a straight
line.
6. TP and TQ are any two tangents to a parabola and the
tangent at a third point B, cuts them in F' and Q' ; prove that
TT' Tq_ QQ' _TP' _Q'R
7. If a parabola touch three given straight lines, prove that each
of the lines joining the points of contact passes through a fixed point.
8. A parabola touches two given straight lines ; if its axis pass
through the point {h, k), the given lines being the axes of coordinates,
prove that the locus of the focus is the curve
x'^-y^-hx + ky=: 0.
9. A parabola touches two given straight lines, which meet at O,
in given points and a variable tangent meets the given lines in P and
Q respectively ; prove that the locus of the centre of the circumcircle
of the triangle OPQ is a fixed straight line.
10. The sides AB and AC of a triangle ABC are given in position
and the harmonic mean between the lengths AB a,nd AC is also given;
prove that the locus of the focus of the parabola touching the sides at
B and C is a circle whose centre lies on the line bisecting the angle
BAG.
11. Parabolas are drawn to touch the axes, which are inclined at
an angle w, and their directrices all pass through a fixed point (h, k).
Prove that all the parabolas touch the straight line
+ ^r-r^ = 1-
h + k sec 0} k + h sec w
CHAPTER XII.
THE ELLIPSE.
247. The ellipse is a conic section in which the
eccentricity e is less than unity.
To find the equation to an ellipse.
Let ZK be the directrix, S the focus, and let SZ be
perpendicular to the directrix.
There will be a point A on SZ, such that
SA = e.AZ (1).
Since e < 1, there will be another point A\ on ZS produced,
such that
SA' = e.A'Z (2).
L. 15
226 COORDINATE GEOMETRY.
Let the length A A! be called 2<a^, and let C be the middle
point of AA', Adding (1) and (2), we have
'la=-AA! = e{AZ^A:Z) = 'l.e.CZ,
i.e. CZ = - (3).
Subtracting (1) from (2), we have
e{A:Z-AZ)^SA'-8A^{SG^CA')-{GA-CS),
i.e. e.AA' = 2CS,
and hence CS — a.e (4).
Let G be the origin, CA' the axis of x, and a line through
C perpendicular to ^^' the axis of 3/.
Let P be any point on the curve, whose coordinates are
X and y, and let Pif be the perpendicular upon the directrix,
and PJV the perpendicular upon A A'.
The focus S is the point (— ae, 0).
The relation SP" = e^ . PM^ = e" . ZN^ then gives
(x + aef-¥y^^e^(x+'^\ (Art. 20),
i.e x%l^e') + y' = a\l-e%
a'{l-e')
If in this equation we put x = Oy we have
y = + a\/l—e^j
shewing that the curve meets the axis of y in two points,
B and B', lying on opposite sides of C, such that
B'C ^CB-^a sjT^\ i.e. CB' = CA' - CS\
Let the length CB be called 6, so that
The equation (5) then becomes
±-+L = l (6).
a2^b2"-^ ^ ^
i-e. -2 + -2^^i — :i^ = l (^)-
THE ELLIPSE. 227
248. The equation (6) of the previous article may be
written
if' Q(? c? — x^ {a + x) {a — x)
Jf a^ c^ o? '
FN'' AN.NA'
i.e. PJV^ : AN.NA' :: BC : AC\
Def. The points A and A' are called the vertices of
the curve, AA' is called the major axis, and BB' the minor
axis. Also C is called the centre.
249. Since ♦S' is the point {—ae, 0), the equation to
the ellipse referred to S as origin is (Art. 128),
The equation referred to A as origin, and AX and a
perpendicular line as axes, is
(^-^)' , ^' _ 1
a- 0^ a
Similarly, the equation referred to ZX and ZK as axes is,
a
since CZ= — ,
(-3'
+ ^- = h
The equation to the ellipse, whose focus and directrix are any
given point and line, and whose eccentricity is known, is easily
written down.
For example, if the focus be the point (-2, 3), the directrix be
the line 2a; + 3?/ + 4 = 0, and the eccentricity be |, the required equa-
tion is
i.e. 261*2 + 181i/2 - 192a;?/ + 1044a; - 2334?/ + 3969 = 0.
15—2
228 COORDINATE GEOMETRY.
Generally, the equation to the ellipse, whose focus is the point
(/> P)> whose directrix is Ax + By + C = 0, and whose eccentricity
is e, is
250. There exist a second focus and a second directrix
for the curve.
On the positive side of the origin take a point S', which
is such that SO — CS' = ae, and another point Z', such that
e
Draw Z'K' perpendicular to ZZ\ and PM' perpen-
dicular to Z'K'.
The equation (5) of Art. 247 may be written in the
form
x^ — 2aex + ah^ + y^ = ^'^^ - 2aeaj + c?,
i.e. {x - aef -^ y^ = e^ (- - x\ ,
i.e. ST^^e^PM"".
Hence any point P of the curve is such that its distance
from S' is e times its distance from Z'K', so that we should
have obtained the same curve, if we had started with S' as
focus, Z'K' as directrix, and the same eccentricity.
251. The sum of the focal distances of any point on the
curve is equal to the major axis.
For (Fig. Art. 247) we have
SP = e.PM, and S'P^e.PM'.
Hence
SP + S'P = e (PM+PM') = e . MM'
= e.ZZ' = 2e.CZ=^2a (Art. 247.)
= the major axis.
Also SP - e . PM^ e.NZ^e.CZ+e.CN-?i + ex',
and S'P - e . PM' = e . NZ' = e . CZ' - e . CN -?i- ex',
where x' is the abscissa of P referred to the centre.
THE ELLIPSE. LATUS-RECTUM. 229
252. Mechanical construction for an ellipse.
By the preceding article we can get a simple mechanical
method of constructing an ellipse.
Take a piece of thread, whose length is the major axis
of the required ellipse, and fasten its ends at the points S
and aS" which are to be the foci.
Let the point of a pencil move on the paper, the point
being always in contact with the string and keeping the
two portions of the string between it and the fixed ends
always tight. If the end of the pencil be moved about on
the paper, so as to satisfy these conditions, it will trace out
the curve on the paper. For the end of the pencil will be
always in such a position that the sum of its distances from
S and S' will be constant.
In practice, it is easier to fasten two drawing pins at S
and aS", and to have an endless piece of string whose total
length is equal to the sum of SS' and AA'. This string
must be passed round the two pins at S and aS" and then be
kept stretched by the pencil as before. By this second
arrangement it will be found that the portions of the curve
near A and A' can be more easily described than in the first
method.
253. Latus-rectum of the ellipse.
Let LSL' be the double ordinate of the curve which
passes through the focus aS'. By the definition of the curve,
the semi-latus-rectum SL
= e times the distance of L from the directrix
^e.SZ=e{CZ-CS) = e.CZ-e.CS
= a — ae^ (by equations (3) and (4) of Art. 247)
= -. (Art. 247.)
254. To trace the curve
%4-' «•
230 COORDINATE GEOMETRr.
The equation may be written in either of the forms
±«yi-s (3).
or X
From (2), it follows that if Q^>a?, i.e. if x> a ov <- a^
then y is impossible. There is therefore no part of the
curve to the right of J.' or to the left of A.
From (3), it follows, similarly, that, if y>h or <:-6,
X is impossible, and hence that there is no part of the curve
above B or below B'.
If X lie between — a and + a^ the equation (2) gives two
equal and opposite values for y^ so that the curve is sym-
metrical with respect to the axis of x.
If y lie between — h and + 5, the equation (3) gives two
equal and opposite values for x^ so that the curve is sym-
metrical with respect to the axis of y.
If a number of values in succession be given to £c, and
the corresponding values of y be determined, we shall
obtain a series of points which will all be found to lie on a
curve of the shape given in the figure of Art. 247.
255. The quantity — ^ + -7^ — 1 i^ negative^ zero, or
€(/
positive, according as the point (x , y') lies vjithin, upon, or
without the ellijjse.
Let Q be the point {x, y'), and let the ordinate QN
tlirough Q meet the curve in P, so that, by equation (6) of
Art. 247,
PiP _ 1 ^'
"W ~ ^ •
If Q be within the curve, then y', i.e. QI^, is < FN, so
that
V'' PN^ . -, x'^
0- h- a^
RADIUS VECTOR IN ANY DIRECTION. 231
Hence, in this case,
I.e. — r + "v^— 1 IS negative.
Similarly, if Q' be without the curve, y' > PN, and then
'2 '2
-^ + ^ - 1 is positive,
a" 6^
256. To find the length of a radius vector from the
centre drawn in a given direction.
The equation (6) of Art. 247 when transferred to polar
coordinates becomes
r^cos^^ r^sin^^
a'W
We thus have the value of the radius vector drawn at any
inclination 6 to the axis.
Since r^ = ^a — r-s — tit^ — r-„-;, , we see that the greatest
62 + (a2-62)sin2(9' *=
value of r is when ^ = 0, and then it is equal to a.
Similarly, B == 90° gives the least value of r, viz. h.
Also, for each value of ^, we have two equal and opposite
values of r, so that any line through the centre meets the
curve in two points equidistant from it.
257. Auxiliary circle. Def. The circle which is
described on the major axis, AA\ of an ellipse as diameter,
is called the auxiliary circle of the ellipse.
Let NP be any ordinate of the ellipse, and let it be
produced to meet the auxiliary circle in Q.
Since the angle AQA' is a right angle, being the angle
in a semicircle, we have, by Euc. vi. 8, QN^ = AN. NA'.
232
COORDINATE GEOMETRY.
SO tliat
Hence Art. 248 gives
PN^ : QN' :: BC^ : AG\
PNBCb
QN""AC"a'
Y
^^^^
Q'
y^^^-^^
P
"^^"^^^
f .
P
t^-...
\ C N' N jA' X 1
The point Q in which the ordinate NF meets the
auxiliary circle is called the corresponding point to P.
The ordinates of any point on the ellipse and the
corresponding point on the auxiliary circle are therefore to
one another in the ratio h : «, i.e. in the ratio of the
semi-minor to the semi-major axis of the ellipse.
The ellipse might therefore have been defined as follows :
Take a circle and from each point of it draw perpen-
diculars upon a diameter ; the locus of the points dividing
these perpendiculars in a given ratio is an ellipse, of which
the given circle is the auxiliary circle.
258. Eccentric Angle. Def. The eccentric angle
of any point F on the ellipse is the angle NCQ made with
the major axis by the straight line CQ joining the centre G
to the point Q on the auxiliary circle which corresponds to
the point P.
This angle is generally called ^.
THE ECCENTRIC ANGLE. 233
We have CJV=CQ . cos, <f> = a cos (f>,
and J^Q = CQ sin (f> = a sin <^.
Hence, by the last article,
a
The coordinates of any point P on the ellipse are there-
fore a cos <f> and b sin ^.
Since F is known when ^ is given, it is often called
" the point <^."
259. To obtain the equation of the straight line joi^iing
two points on the ellij^se whose eccentric angles are given.
Let the eccentric angles of the two points, P and P\ be
ff> and ^', so that the points have as coordinates
{a cos ^, b sin ^) and (a cos <^', 6 sin ^f)').
The equation of the straight line joining them is
, . , 6 sin d)' - 6 sin </> ,
2/ - 6 sm <h -, V (x — a cos d))
a cos fp —a cos ^ ^ '
_b 2cosl(<^ + <^>inl(</>--c^)
a 'Ssin !(<}!> + </,') sin i(<^-</>y'^ «cos<^;
6 cosl(<^ + <^') . .
= - - . -i— f-h , — 7\ {x-a cos </>),
a ' sin J (^' + <^)
i.e.
a; (ft + cb' y . ch + ch' 4> + ^' . .6 + 6'
- cos — ~ — + Y- Sin -^~ - = cos <i> cos ^ + sin </> sm ^
C&262 2 ^2
r «^ + </,'-l <^-c^' ,-,
-cos|^<^--^^-^J=:cos^^l-^ (1).
This is the required equation.
Cor. The points on the auxiliary circle, corresponding to P and
P', have as coordinates (a cos 0, a sin 0) and (a cos 0', a sin 0')-
The equation to the line joining them is therefore (Art. 178)
X <t> + <i) V . (i> + (h' (i>-<t>'
- cos ^— -^ + •- sin --^ =cos —r- •
a 2 a 2 2
234 COORDINATE GEOMETRY.
This straight line and (1) clearly make the same intercept on the
major axis.
Hence the straight line joining any two points on an ellipse, and
the straight line joining the corresponding points on the auxiliary
circle, meet the major axis in the same point.
EXAMPLES. XXXII.
1. Find the equation to the ellipses, whose centres are the
origin, whose axes are the axes of coordinates, and which pass
through (a) the points (2, 2), and (3, 1),
and (/3) the points (1, 4) and (-6, 1).
Find the equation of the ellipse referred to its centre
2. whose latus rectum is 5 and whose eccentricity is |,
3. whose minor axis is equal to the distance between the foci and
whose latus rectum is 10,
4. whose foci are the points (4, 0) and ( - 4, 0) and whose
eccentricity is ^.
5. Find the latus rectum, the eccentricity, and the coordinates
of the foci, of the ellipses
(1) a:^ + Sy^ = a^, (2) 5x'^ + 4y'^ = l, and (3) 9x^ + 5y^-S0y = O.
6. Find the eccentricity of an ellipse, if its latus rectum be equal
to one half its minor axis.
7. Find the equation to the ellipse, whose focus is the point
(-1, 1), whose directrix is the straight line x -y + S=0, and whose
eccentricity is ^.
8. Is the point (4, - 3) within or without the ellipse
5x^ + 7y^ = lV?
9. Find the lengths of, and the equations to, the focal radii drawn
to the point (4 ^3, 5) of the ellipse
25x2 + 162/2=1600.
10. Prove that the sum of the squares of the reciprocals of two
perpendicular diameters of an ellipse is constant.
11. Find the inclination to the major axis of the diameter of the
ellipse the square of whose length is (1) the arithmetical mean,
(2) the geometrical mean, and (3) the harmonical mean, between the
squares on the major and minor axes.
12. Find the locus of the middle points of chords of an ellipse
which are drawn through the positive end of the minor axis.
13. Prove that the locus of the intersection of AP with the
straight line through A' perpendicular to A'F is a straight line which
is perpendicular to the major axis.
[EXS. XXXII.] THE ECCENTRIC ANGLE. 235
14. Q is the point on the auxiliary circle corresponding to P on
the ellipse; PLM is drawn parallel to CQ to meet the axes inL andilf ;
prove that PL = b and PM—a.
15. Prove that the area of the triangle formed by three points on
an ellipse, whose eccentric angles are ^, 0, and \{/, is
. , . 0-^ . yl/-d . d-(t>
lab sin -^-^ sm i-^— sm — ^ .
Prove also that its area is to the area of the triangle formed by the
corresponding points on the auxiliary circle as 6 : a, and hence that
its area is a maximum when the latter triangle is equilateral, i.e. when
27r
0-^ = ^-0 = —.
16. Any point P of an ellipse is joined to the extremities of the
major axis; prove that the portion of a directrix intercepted by them
subtends a right angle at the corresponding focus.
17. Shew that the perpendiculars from the centre upon all chords,
which join the ends of perpendicular diameters, are of constant
length.
18. If a, j8, 7, and 5 be the eccentric angles of the four points of
intersection of the ellipse and any circle, prove that
a + /3 + 7 + 5isan odd multiple
of IT radians.
[See Trigonometry, Part II, Art. 31, and Page 37, Ex. 15.]
19. The tangent at any point P of a circle meets the tangent at a
fixed point A in T, and T is joined to B, the other end of the
diameter through A ; prove that the locus of the intersection of AP
and BT is an ellipse whose eccentricity is — .- .
20. From any point P on the ellipse, PN is drawn perpendicular
to the axis and produced to Q, so that NQ equals PS, where ^ is a
focus ; prove that the locus of Q is the two straight lines y±ex + a = 0.
21. Given the base of a triangle and the sum of its sides, prove
that the locus of the centre of its incircle is an ellipse.
22. With a given point and line as focus and directrix, a series
of ellipses are described; prove that the locus of the extremities of
their minor axes is a parabola.
23. A line of fixed length a + b moves so that its ends are always
on two fixed perpendicular straight lines; prove that the locus of a
point, which divides this line into portions of length a and b, is an
ellipse.
24. Prove that the extremities of the latera recta of all ellipses,
having a given major axis 2a, lie on the parabola x^= -a{y- a).
236 COORDINATE GEOMETRY,
260. Tojind the intersections of any straight li7ie with
the ellipse
T + |I=1 (1).
Let the equation of the straight line be
y ^ TThx + c (2).
The coordinates of the points of intersection of (1) and
(2) satisfy both equations and are therefore obtained by
solving them as simultaneous equations.
Substituting for y in (1) from (2), the abscissae of the
points of intersection are given by the equation
^ (mx + cf
a' '^ ^ir~ " '
i.e. x" {a^mP + b') + 2a^mcx + a' (c" - If) = (3).
This is a quadratic equation and hence has two roots,
realj coincident, or imaginary.
Also corresponding to each value of x we have from (2)
one value of y.
The straight line therefore meets the curve in two points
real, coincident, or imaginary.
The roots of the equation (3) are real, coincident, or
imaginary according as
(2a^?7ic)^— 4 (b^+a^ni^) x a^ (d^—lr) is positive, zero, or negative,
i.e. according as h^(h'^-\-a^m?)—l?c^ is positive, zero, or negative,
i.e. according as c^ is < =: or > ahn^ + h^.
261. To find the length of the chord intercepted hy the
ellipse on the straight line y = mx + c.
As in Art. 204, we have
Marine , or {c^ — Ir)
X-, -f- x^ ^= — ~ 5 Yh J ^nci X1X2 =^ s 5 7 „ ,
, , , 2a6 sja-nr + b' — c
so that X, — Xo = 7,-7, — Th
a^mr + o"
EQUATION TO THE TANGENT 237
The length of the required chord therefore
= J{x^ - flJa)^ + (2/1 - 2/2)^ = {^1 -x^ sll+m^
2ab \/l + m^ Ja^m^ + b^ — c^
a^m^ + b^
262. To find the equation to the tangent at any 'point
{x\ y') of the ellipse.
Let P and Q be two points on the ellipse, whose coordi-
nates are {x\ y') and {x", y").
The equation to the straight line PQ is
y-2/'=J^|'(^-*') (!)•
Since both P and Q lie on the ellipse, we have
10 ro
a-'¥=' • (2)'
,2
-^^+v=l (^>-
Hence, by subtraction,
x"'-x'^ y"^-y''_^
a^ "^ 6^ ~ '
( y"-y){y" + y') _ _ { x"-x'){x" + x')
y"-y' _b^ x" + x'
**^' ^^'^^''" a''y" + y''
On substituting in (1) the equation to any secant PQ
becomes
f 0'' X -T X . ,v J .^
y-y- 5-77 ,{x~x) (4).
To obtain the equation to the tangent we take Q
indefinitely close to P, and hence, in the limit, we put
a;" = X and y" = y ,
238 COORDINATE GEOMETRY.
The equation (4) then becomes
/ ox, •>
a" 0" c? h^
i.e. -2 + -.T = -2 + y^ = 1' % equation (2).
The required equation is therefore
a2 "*■ b2 " ■*■•
Cor. The equation to the tangent is therefore ob-
tained from the equation to the curve by the rule of
Art. 152.
263. To find the equation to a tangent in terms of the
tcmgent of its inclination to the major axis.
As in Art. 260, the straight line
y = 7nx + c (1)
meets the ellipse in points whose abscissae are given by
af (62 + ahn") + Imca'x + a" (c^ - h'') = 0,
and, by the same article, the roots of this equation are
coincident if
c = \ja?m^ + y^.
In this case the straight line (1) is a tangent, and
it becomes
y = mx+ Va2m2 + b2 (2).
This is the required equation.
Since the radical sign on the right-hand of (2) may
have either + or — prefixed to it, we see that there are two
tangents to the ellipse having the same w^, i.e. there are
two tangents parallel to any given direction.
The above form of the equation to the tangent may be deduced
from the equation of Art. 262, as in the ease of the parabola
(Art. 206). It will be found that the point of contact is the point
/ - c?'m y^ \
EQUATION TO THE TANGENT. 239
264. By a proof similar to that of the last article, it
may be shewn that the straight line
X cos a + y sin a =^p
touches the ellipse, if
p2 = a2 cos2 a + b2 sin2 a.
Similarly, it may be shewn that the straight line
Ix + my = n
touches the ellipse, if aH^ + b^m^ = v?.
265. Equation to the tangent at the point whose
eccentric angle is <f>.
The coordinates of the point are (a cos ^, b sin (ft).
Substituting x' = a cos <^ and y' — b sin <f> in the equation
of Art. 262, we have, as the required equation,
X V
- COS ^ + ^ sin ^ = 1 (1).
3> D
This equation may also be deduced from Art. 259.
For the equation of the tangent at the point "<^" is
obtained by making cf>' = (ft in the result of that article.
Ex. Find the intersection of the tangents at the points <p and <p'.
The equations to the tangents are
- COS + V sin - 1 = 0,
a
and - cos 0' + ^ sin 0' - 1 = 0.
a b
The required point is found by solving these equations.
We obtain
X y '
a b -1 1
sin - sin 0' cos <p' - cos <p sin 0' cos - cos <p' sin sin (0 - 0') '
i.e.
5 y 1
(b + <b . (b — d)' ^, . + 0'. (b — <b ^ , (h — d) dt — <b
2a COS ^ sin ^ ^ 2b sin ^ ■ sm - 2 sin ~^ cos ■ ^
240 COORDINATE GEOMETRY.
Hence ^=a '=-2iiM±£) and j, = i,!i^i <|±J3.
COS i{(f>-^') COS J (0 - <p')
266. Equation to the normal at the point {x, y').
The required normal is the straight line which passes
through the point {x\ y) and is perpendicular to the
tangent, i.e. to the straight line
a'y y
Its equation is therefore
y — y' —m(x — x')^
where m ( — ^-\ = -l, i.e. m ^ ^, , (Art. 69).
\ ay / o-'x
ft u
The equation to the normal is therefore y — y' = -— -, {x - x),
X - X' y - y
^.e.
a2 b2
267. Equation to the normal at the point whose eccentric
angle is <j>.
The coordinates of the point are a cos <^ and b sin (f>.
Hence, in the result of the last article putting
X —a cos ^ and y =h sin <j!>,
. , , x — a cos c& y — h sin <i>
it becomes , =: . , ,
cos <p sm <^
a h
ax „ by
cos <p sm ip
The required normal is therefore
ax sec ^ — by cosec ^ = a2 — b2.
SUBTANGENT AND SUBNORMAL.
# 268. Equation to the normal in the form y = mx + c.
The equation to the normal at {x', y') is, as in Art. 266,
241
_ , a^u ^- ^ X ay
Let^, = m,sothat- = ^-^^.
Hence, since {x', y') satisfies the relation -^ + ^ = 1> ^^^ obtain
y"^
62
b^m
y ~ Ja^ + hHi^'
The equation to the normal is therefore
y=mx- f- ^.
This is not as important an equation as the corresponding equa-
tion in the case of the parabola. (Art. 208.)
When it is desired to have the equation to the normal expressed
in terms of one independent parameter it is generally better to use
the equation of the previous article.
269. To find the length of the suhtangent and sub-
normal.
Let the tangent and normal at P, the point (x', 3/'),
meet the axis in T and G respectively, and let PN be the
ordinate of P,
L.
16
242 COORDINATE GEOMETRY.
The equation to the tangent at P is (Art. 262)
? + f^=l (!)•
To find where the straight line meets the axis we put
y — and have
x=-, i.e. CT^
^ ' CN'-
I.e.
GT.GN=a^=GA^ (2).
Hence the subtangent NT
a? , a^ — x'^
^CT-GN=^-x'^
X X
The equation to the normal is (Art. 266)
X — X y ~y
x y' '
To find where it meets the axis, we put 2/ — 0, and have
X X y 7 ■>
^^^^~ / ~ ^ J
x_ ^
'a? ¥
i.e. Ga=^x = x'-^,x'='^^x'=^eKx'^e\GN...{3).
CO CO
Hence the subnormal JVG
^GN-GG^{l-e^)GN,
i.e. NG::NGv.l~e^'.l
:: h" : a\ (Art. 247.)
Cor. If the tangent meet the minor axis in t and Pn
be perpendicular to it, we may, similarly, prove that
Gt.Gn^h\
270. Some properties of the ellipse.
(a) SG = e,SP, and the tangent and normal at P bisect the
external and internal angles between the focal distances of P.
By Art. 269, we have CG=e^x'.
SOME PROPERTIES OF THE ELLIPSE. 243
Hence SG = SC+CG = ae + e^x' = e.SP, by Art. 251.
Also S'G=zCS'-CG = e{a-ex') = e.S'P.
Hence SG : S'G :: SP : S'P.
Therefore, by Euc. vi, 3, PG bisects the angle SPS'.
It follows that the tangent bisects the exterior angle between
SP and ST.
(/3) If SY and S'Y' be the perpendiculars from the foci upon the
tangent at any point P of the ellipse, then Y and Y' lie on the auxiliary
circle, and SY . S'Y' = b'^. Also GY and S'P are parallel.
The equation to any tangent is
X co^a + y Bva.a=p (1),
where p = sja^ cos^ a + 6^ gin^ a (Art. 264).
The perpendicular SY to (1) passes through the point {-ae, 0)
and its equation, by Art. 70, is therefore
(a; + ae)sin a- j/cos a = (2).
If Y be the point {h, k) then, since Y lies on both (1) and (2), we
have
h cos a + k sin a — sja^ cos^ a + b^ sin^ a,
and h sin a - ^ cos a= -ae sin a = - ija^ - b^ sin^ a.
Squaring and adding these equations, we have h'^ + k'^ = a'^, so that
Y lies on the auxihary circle x^ + y'^:= c^.
Similarly it may be proved that Y' lies on this circle.
Again S is the point ( - ae, 0) and S' is (ae, 0).
Hence, from (1),
SY=p-\-ae cos a, and S'Y'=p - ae cos a. (Art. 75.)
Thus SY . S'Y' =^2 _ a2g2 cos2 a
= a^ cos^ a + &2 sin^ a-{a'^- b^) cos^ a
Also CT= "'
GN'
a^ _a{a-eCN)
and therefore S'T=~-^-ae= ^,,
ON CN
GT «^__CT
''■ S'T~ a-e.G~N~ S'P'
Hence GY and S'P are parallel. Similarly GY' and SF are
parallel.
16—2
244 COORDINATE GEOMETRY.
(7) If the normal at any point P meet the major and minor axes
in G and g, and if GF he the perpendicular upon this normal, then
FF.PG^ &2 aj^^ PF.Pg = a2.
The tangent at any point P (the point " 0") is
ah
Hence Pi^= perpendicular from G upon this tangent
1 ah
V
cos^ sin^ (p fjb^ cos^ + a'-^ sin^
(1).
The normal at P is
-^--X=a^-h- (2).
cos sm
If we put y = 0, we have GG = cos 0.
a cos - cos J + 62 sin3
h^
= "2 COS^ + &2 sin^ 0,
i.e. PG^-Jb^cos^(p + a^sin^(p.
a ^
From this and (1), we have PF.PG = h^.
If we put a; = in (2), we see that g is the point
(0, ^sin0j.
Hence Pg^ = a^cos^(p+i h sin 04 ^ — sin j ,
so that ^^~h '^^^ ^^^^ + ^2 sin^ 0.
From this result and (1) we therefore have
PF. Pg = a\
271. To find the locus of the point of intersectio7i of
tangents which meet at right angles.
Any tangent to the ellipse is
y — 7nx + sja^rri^ + 6'^,
and a perpendicular tangent is
y--in''^\J'''{-^3*^^-
TANGENTS AND NORMALS. EXAMPLES. 245
Hence, if {h, k) be their point of intersection, we have
k — mil — sjahii? + fe- (1),
and ink + A = sjo? + Ir'nn? (2).
If between (1) and (2) we eliminate m, we shall have a
relation between h and k. Squaring and adding these
equations, we have
(P + W) (1 + m^) = (^2 + W) (1 + m%
i.e. h'' + k^ = a^ + h\
Hence the locus of the point (Ji, k) is the circle
a^ + 2/^ = «- + h^,
i.e. a circle, whose centre is the centre of the ellipse, and
whose radius is the length of the line joining the ends
of the major and minor axis. This circle is called the
Director Circle.
EXAMPLES. XXXIII.
Find the equation to the tangent and normal
1. at the point (1, f) of the ellipse 4a;2 + Qt/^ = 20,
2. at the point of the ellipse 5a;2 + 3z/2 = 137 whose ordinate is 2,
3. at the ends of the latera recta of the ellipse ^x^ + 16?/2 = 144.
4. Prove that the straight line y = x + ^^^ touches the ellipse
5. Find the equations to the tangents to the ellipse 4j,'- + 3?/^ = 5
which are parallel to the straight line y = Sx + 7.
Find also the coordinates of the points of contact of the tangents
which are inclined at 60° to the axis of x.
6. Find the equations to the tangents at the ends of the latera
recta of the ellipse -g + T2 =^ •'■» ^^^ shew that they pass through the
intersections of the axis and the directrices.
7. Find the points on the ellipse such that the tangents there
are equally inclined to the axes. Prove also that the length of the
perpendicular from the centre on either of these tangents is
2 '
x/'
246 COORDINATE GEOMETRY. [ExS.
8. In an ellipse, referred to its centre, the length of the sub-
tangent corresponding to the point (3, V) is -V-; prove that the
eccentricity is f .
9. Prove that the sum of the squares of the perpendiculars on
any tangent from two points on the minor axis, each distant jja?' - fe^
from the centre, is ^a?-.
10. Find the equations to the normals at the ends of the latera
recta, and prove that each passes through an end of the minor axis if
e4 + e2 = l.
11. If any ordinate MP meet the tangent at .L in Q, prove that
MQ and SP are equal.
12. Two tangents to the ellipse intersect at right angles; prove
that the sum of the squares of the chords which the auxiliary circle
intercepts on them is constant, and equal to the square on the line
joining the foci.
13. If P be a point on the ellipse, whose ordinate is y', prove
that the angle between the tangent at P and the focal distance of P
is tan~i — ; .
aey
14. Shew that the angle between the tangents to the ellipse
— \.^ = '\. and the circle x~ + y'^ = ab at their points of intersection is
a^ Ir
a— b
tan~^ ~7=? .
y/ab
15. A circle, of radius r, is concentric with the ellipse ; prove
that the common tangent is inclined to the major axis at an angle
tan~i X I —, H and find its length,
16. Prove that the common tangent of the ellipses
a;2 y'^ _'ix T x^ y^ 2x _
a^ b^ c />- a'^ c
subtends a right angle at the origin.
17. Prove that PG.Pg = SP. ST, and CG.CT= CS^.
18. The tangent at P meets the axes in T and t, and CY is the
perpendicular on it from the centre; prove that (1) Tt . PY=a--b'^,
and (2) the least value oiTtisa + b.
19. Prove that the perpendicular from the focus upon any tangent
and the line joining the centre to the point of contact meet on the
corresponding directrix.
20. Prove that the straight lines, joining each focus to the foot of
the perpendicular from the other focus upon the tangent at any
point P, meet on the normal at P and bisect it.
XXXIIIJ TANGENTS AND NOKMALS. EXAMPLES. 247
21. Prove that the circle on any focal distance as diameter touches
the auxiliary circle.
22. Find the tangent of the angle between CP and the normal at
P, and prove that its greatest value is ^ , .
2a6
23. Prove that the straight line lx + my=n is a normal to the
a^ b'~ (a^ - 6^)2
fillipse, if i2 + — 2 = 9- •
24. Find the locus of the point of intersection of the two straight
lines - - l + t = and - + ^ - 1 = 0.
a a
Prove also that they meet at the point whose eccentric angle is
2tan-ii.
25. Prove that the locus of the middle points of the portions of
tangents included between the axes is the curve
26. Any ordinate NP of an ellipse meets the auxiliary circle in
Q ; prove that the locus of the intersection of the normals at P and
Q is the circle x^ + y^ = {a + h) ^.
27. The normal at P meets the axes in G and g ; shew that the
loci of the middle points of PG and Gg are respectively the ellipses
4^.2 4^,2
^2(1^,2)2 + |- = 1, anda%2 + 62^2^|(,,2_62)2.
28. Prove that the locus of the feet of the perpendicular drawn
from the centre upon any tangent to the ellipse is
r2 = a2 cos2 ^ + &2 sin2 e. [ Use Art. 264.]
29. If a number of ellipses be described, having the same major
axis, but a variable minor axis, prove that the tangents at the ends of
their latera recta pass through one or other of two fixed points.
30. The normal GP is produced to Q, so that GQ = n. GP.
Prove that the locus of Q is the ellipse -rn » jtck + -^o = 1.
a^{n + e^-ne^y n%^
31. If the straight line y = mx + c meet the ellipse, prove that the
equation to the circle, described on the line joining the points of
intersection as diameter, is
(a2m2 + 62) (a;2 + y2) + 2ma^cx - 2b'^cij + c^ {a^ + 62) - a262 (1 + wF) = 0.
32. PM and PN are perpendiculars upon the axes from any point
P on the ellipse. Prove that MN is always normal to a fixed
concentric ellipse.
248 COORDINATE GEOMETRY. [EXS. XXXIII.]
33. Prove that the sum of the eccentric angles of the extremities
of a chord, which is drawn in a given direction, is constant, and
equal to twice the eccentric angle of the point at which the tangent is
parallel to the given direction.
34. A tangent to the ellipse ~2 + fa — -^ meets the ellipse
in the points P and Q; prove that the tangents at P and Q are at
right angles.
272. To prove that through any given point (cci, 3/1)
there pass, in general, two tangents to an ellipse.
The equation to any tangent is (by Art. 263)
y = mx + sja^m^ + b^ (1).
If this pass through the fixed point (x-^^, y-^, we have
2/1 — mcci = sjo^m? + IP-,
i.e. y^ — Imx-^y^ + w^x^ — c^w? + IP",
i.e. m^{x^-a^)-'imXjy^+{y{--¥)^Q (2).
For any given values of x^ and y^ this equation is in
general a quadratic equation and gives two values of m
(real or imaginary).
Corresponding to each value of m we have, by sub-
stituting in (1), a different tangent.
The roots of (2) are real and different, if
(- 2x^y^y - 4 (a?!^ — a^) {y^ - If) be positive,
i. e. if IPx^ + a^y^ — a^lP be positive,
X " 11
i.e. ii ^ + ^ - 1 be positive,
a^ 0^
i.e. if the point {x^, y-^ be outside the curve.
The roots are equal, if
h-x^^ + a'^y^^ - a-62
be zero, i.e. if the point {x^, y-^ lie on the curve.
CHORD OF CONTACT OF TANGENTS. 249
The roots are imaginary, if
be negative, i.e. if the point (a?i, y^ lie within the curve
(Art. 255).
273. Equation to the chord of contact of tangents
drawn from a point {x^, y-^).
The equation to the tangent at any point Q, whose
coordinates are x and y', is
^ + ^' =, 1
Also the tangent at the point R, whose coordinates are
x" and y\ is
H It
a^ Jy"
If these tangents meet at the point 57, whose coordi-
nates are x^ and y^, we have
a^ 62 -^ l^^
and ^' + 2/_|:^l ^2).
a^ 0^ ^ ^
The equation to ^^ is then
^^^=^ (3)-
For, since (1) is true, the point {x, y') lies on (3).
Also, since (2) is true, the point {x'\ y") lies on (3).
Hence (3) must be the equation to the straight line
joining (x', y) and {x\ y"), i.e. it must be the equation to
QR the required chord of contact of tangents from {x^, v/i)-
274. To find the equation of the polar of the point
(•''^1) 2/i) '^'^'i^t' Tespect to the ellip)se
S-J=l- [Art. 162.]
250 COORDINATE GEOMETRY.
Let Q and M be the points in which any chord drawn
through the point (ccj, y-^ meets the ellipse [Fig. Art. 214].
Let the tangents at Q and R meet in the point whose
coordinates are (A, h).
"We require the locus of {Ji, k).
Since QR is the chord of contact of tangents from
(A, ^), its equation (Art. 273) is
xh yk
~^'^'¥^
Since this straight line passes through the point (x^, ^j),
we have
hx^ ky^
a
+ ^ = 1 (!)•
Since the relation (1) is true, it follows that the point
(A, k) lies on the straight line
^^W=^ (^)-
Hence (2) is the equation to the polar of the point
Cor. The polar of the focus (ae, o) is
X. ae ^ . a
a^ e
i.e. the corresponding directrix.
275. "When the point (iCj, y^ lies outside the ellipse,
the equation to its polar is the same as the equation of the
chord of contact of tangents from it.
When (a^i, y^ is on the ellipse, its polar is the same as
the tangent at it.
As in Art. 215 the polar of (x^^ y^ might have been
defined as the chord of contact of the tangents, real or
imaginary, drawn from it.
276. By a proof similar to that of Art. 217 it can be
shewn that If the polar of P pass through T, then the polar
of T passes through P.
PAIR OF TANGENTS FROM ANY POINT. 251
277. To find the coordinates of the pole of any given
Ixii/e
Ax + By + C=0 (1).
Let (x^, 2/i) be its pole. Then (1) must be the same as
the polar of (a^i, y^)j i.e.
^" + f'-l=0 ..(2).
Comparing (1) and (2), as in Art. 218, the required pole
is easily seen to be
278. To find the equation to the pair of tangents that
can be drawn to the ellipse from the point (x^^j y^.
Let (h, k) be any point on either of the tangents that
can be drawn to the ellipse.
The equation of the straight line joining (A, k) to
(fljj, 2/i) is
^ - Vl / x
k — y^ hy. — kx.
t.e. y = j — ^x+^ ^\
If this straight line touch the ellipse, it must be of the
form
y = mx + s]a^w? + h^. ( Art. 263.)
Hence
rnJ^, and f^^lJl^V = „W + Jl
h — x^ \ h — x^ J
Hence f ^^' " ^■ Y = «^ faV + 6'.
\ h-x^ J \h-xj
But this is the condition that the point (A, U) may lie
on the locus
{xy^ - x^yf =a'{y- y^f + b''{x- x^f (1).
This equation is therefore the equation to the required
tangents.
252 COORDINATE GEOMETRY.
It would be found that (1) is equivalent to
/^' ,2/' i\ /^^i' ^3// A _ f^^i ^ 2/2/1 1
279. To Jind the locus of the middle points of parallel
chords of the ellij)se.
Let the chords make with the axis an angle whose
tangent is m, so that the equation to any one of them,
QR, is
y = 7nx + c (1),
where c is different for the different chords.
This straight line meets the ellipse in points whose
abscissae are given by the equation
x^ {mx + cY
a-
-1.
i. e. ^ {a'm' + 6') + la'mcx + a'{c'-b^) = (2).
Let the roots of this equation, i.e. the abscissae of Q
and H, be x^ and x^, and let F, the middle point of QB, be
the point [h, k).
Then, by Arts. 22 and 1, we have
tJC-i *T* t^o
a^mc
C^Vl^ + IP'
(3).
CONJUGATE DIAMETERS. 253
Also V lies on the straight line (1), so that
k = mh + c (4).
If between (3) and (4) we eliminate c, we have
. a^TTb (k — tnh)
^" a?rri' + }f '
i. e. hVi = ~ a^mk (5).
Hence the point (A, k) always lies on the straight line
y — — 5 — X (6).
The required locus is therefore the straight line
y — 7n-,x, where m, = ^r— ,
^'
I.e. mm. — -„ (7).
280. Equation to the chord ivhose middle point is {h, k).
The required equation is (1) of the foregoing article, where m and
c are given by equations (4) and (5), so that
b^h , a?k^ + y-h'^
m=-^.andc = __^-.
The required equation is therefore
^~ a'k''^ a?k '
i.e. ^^{y-Jc)+~{X-h)=0.
It is therefore parallel to the polar of {h, k).
281. Diameter. Def. The locus of the middle
points of parallel chords of an ellipse is called a diameter,
and the chords are called its double ordinates.
By equation (6) of Art. 279 we see that any diameter
passes through the centre C.
Also, by equation (7), we see that the diameter 7/ — m^x
bisects all chords parallel to the diameter y — mx, if
62
mm
'-a^ W-
254 COORDINATE GEOMETRY.
But the symmetry of the result (1) shows that, in this
case, the diameter y — rax bisects all chords parallel to the
diameter y — m^x.
Such a pair of diameters are called Conjugate Diameters.
Hence
Conjugate Diameters. Def. Two diameters are
said to be conjugate when each bisects all chords parallel
to the other.
Two diameters y — tnx and y — m^x are therefore con-
jugate, if
miXli = — — g- ,
^ a-*
282. The tangent at the extremity of any diameter is
parallel to the chords which it bisects.
In the Figure of Art. 279 let [x, y') be the point P on
the ellipse, the tangent at which is parallel to the chord
QBj whose equation is
y — inx + c (1).
The tangent at the point (a;', y) is
^' + ^'=1 (2).
a-' o" ^ '
Since (1) and (2) are parallel, we have
a-'y
i.e. the point (cc', y^ lies on the straight line
a^m,
But, by Art. 279, this is the diameter which bisects QR
and all chords which are parallel to it.
Cor. It follows that two conjugate diameters CP and
CD are such that each is parallel to the tangent at the
extremity of the other. Hence, given either of these, we
have a geometrical construction for the other.
CONJUGATE DIAMETERS. 255
283. The tangents at the ends of any chord meet on the
diameter which bisects the chord.
Let the equation to the chord QR (Art. 279) be
2/ = ma;+ c (1).
Let T be the point of intersection of the tangents at Q
and Rj and let its coordinates be x^ and y^.
Since QR is the chord of contact of tangents from T^ its
equation is, by Art. 273,
^V^^l (2)
The equations (1) and (2) therefore represent the same
straight line, so that
})%
i.e. {h^ k) lies on the straight line
am,
which, by Art. 279, is the equation to the diameter bisect-
ing the chord QR. Hence T lies on the straight line GP.
284. If the eccentric angles of the ends, P and D, of a
pair of conjugate diameters he <^ and (f>', then <^ and cf>' differ
by a right angle.
Since P is the point [a cos ^, h sin </>), the equation to
CPis
y~x.— tan </> (1).
y— X . - tan <^' (2).
a
So the equation to CD is
h
a
These diameters are (Art. 281) conjugate if
6^ , ¥
— tan d> tan ch ~ - — ^ ,
a^ ^ ^ a^
i. e. if tan <^ ^^ — cot cf}' — tan (<^' ± 90°j,
i.e. if ^-^'= + 90°.
256
COOEDINATE GEOMETRY.
Cor. 1. The points on the auxiliary circle correspond-
ing to P and D subtend a right angle at the centre.
For if p and d be these points then, by Art. 258, we
have
z.79(7^'-<^ and LdCA'^^'.
Hence
LpGd=^ LdCA' - LpGA' = <f>-<j>' = 90\
Cor. 2. In the figure of Art. 286 if P be the point <fi,
then D is the point ^ + 90° and D' is the point <^ - 90°.
285. From the previous article it follows that if P be
the point {a cos c(>, b sin <^), then D is the point
{a cos (90° + <^), b sin (90° + <^)} i- e. (— a sin <^, b cos </>).
Hence, if PJ^ and DM be the ordinates of P and i),
we have
iTP CM
and
CJSr MD
a
a
286. If PCP' and BCD' be a pair of conjugate dia-
meierSj then (1) CP^ + CD^ is constant, and (2) the area of
the parallelogram formed by the tangents at the ends of these
diameters is constant.
CONJUGATE DIAMETERS. 257
Let P be the point ^, so that its coordinates are a cos ^
and h sin <^. Then D is the point 90° + </>, so that its co-
ordinates are
a cos (90° + ^) and h sin (90°+ <^),
i.e. —a sin ^ and h cos <^.
(1) We therefore have
CP^ = «' cos^ <^ + 6' sin^ <^,
and CD'^ - a? sin^ cj> + b"^ cos^ ^,
Hence CP^ + CB' = 0"+ b'^
— the sum of the squares of the semi-axes of the ellipse.
(2) Let KLMN be the parallelogram formed by the
tangents at P, D, P', and D'.
By Euc. I. 36j we have
area KLMN = 4 . area CPXD
-4. GU.PK=iCU.CD,
where CU is the perpendicular from C upon the tangent
at P.
Now the equation to the tangent at P is
- cos <l> + T sin (i) — 1 = 0,
a
so that (Art. 75) we have
1 ab ab
C (J
/cos^
^</) sin2<^ J a" sin'' <f, + b^ cos' <f> CD
Hence CU.CI) = ab.
Thus the area of the parallelogram KLMN = iab,
which is equal to the rectangle formed by the tangents
at the ends of the major and minor axes.
287. The product of the focal distances of a point P is
equal to the square on the semidiameter parallel to the tangent
at P.
If P be the point (p, then, by Art. 251, we have
SP = a + ae cos tf), and S'P —a — ae cos ^.
L. 17
258 COORDINATE GEOMETRY.
Hence SP . S'P ^a'~ are' cos' <^
= a^ sin^ <^+b' cos^ <^
288. Ex. If P and D be the ends of conjugate diameters, find
the locus of
(1) the middle point of PD,
(2) the intersection of the tangents at P and D,
and (3) the foot of the perpendicular from the centre upon PD.
P is the point (a cos 0, h sin 0) and D is ( - a sin 0, 6 cos 0).
(1) If (x, y) be the middle point of PD, we have
a cos d>-a sin d) ^ h sin + & cos
^= ^ ^' ^^^ ^ = 2 •
If we eliminate we shall get the required locus. We obtain
2 2
^ + |2=i[(cos - sin 0)2+ (sin + cos 0)2] = ^.
The locus is therefore a concentric and similar ellipse.
[N.B. Two ellipses are similar if the ratios of their axes are the
same, so that they have the same eccentricity.]
(2) The tangents are
-cos0 + vsin0=l,
a b
and -- sin0 + rcos0 = l.
a b ^
Both of these equations hold at the intersection of the tangents.
If we eliminate we shall have the equation of the locus of their
intersections.
By squaring and adding, we have
so that the locus is another similar and concentric ellipse.
(3) By Art. 259, on putting 0' = 9O° + 0, the equation to PD is
- cos (45° + 0) + 1 sin (45° + 0) = cos 45°.
Let the length of the perpendicular from the centre be 2^ and let it
make an angle w with the axis. Then this line must be equivalent to
xcosu + ysm(a=p.
CONJUGATE DIAMETERS. 259
Comparing the equations, we have
..^r. X a cos CO cos 45° , . , , _„ , . & sin w cos 45°
cos (45° + ^) = , and sin(4o° + <^) =
Hence, by squaring and adding, 2p2=a2cos2 w + ft^sin^w, i.e. the
locus required is the curve
289. Equiconjugate diameters. Let P and D be ex-
tremities of equiconjugate diameters, so that CP^ = CDK
If the eccentric angle of P be 0, we then have
a^ cos^ (f> + b^ sin^ (f> = a^ sin^ (f} + h^ cos^ ^,
giving tan^ ^ = Ij
i.e. ^ = 45°, or 135°.
The equation to CP is then
h
y = x. — tan ^,
*.e. 2/ = ±-^ (1)>
and that to CD is y~ — x — cot <^,
a
^.e.
2/ = + -^ (2).
If a rectangle be formed whose sides are the tangents
at Aj A', £, and B' the lines (1) and (2) are easily seen to
be its diagonals.
The directions of the equiconjugates are therefore along
the diagonals of the circumscribing rectangle.
The length of each equiconjugate is, by Art. 286,
290. Supplemental chords. Def. The chords
joining any point P on an ellipse to the extremities, P and
P'j of any diameter of the ellipse are called supplemental
chords.
Supplemental chords are parallel to conjugate diatneters.
17—2
260 COORDINATE GEOMETRY.
Let P be the point whose eccentric angle is <f>, and JR
and E' the points whose eccentric angles are <^i and
180° + <^i.
The equations to FB and FE' are then (Art. 259)
^ oo^ *^ + ^^ + ^ sin ilii-cos ^^1 a^
and
X ^+180° + ^! 2/ . ^+180°+^_ <^_180°-^
— cos ;r -r ^ Sm ^;; — • COS -z:
a 2 b 2 2 '
*.e. - - sm ^Hr^ + y cos ^ ^ ^^ = sm ^ ^ ^\ .. (2).
a 2 6 2 2 ^ '
The « m " of the straight line (1) = cot t±il .
The " m " of the line (2) = - tan ^^^ .
CO A
If
The product of these " m's " = g , so that, by Art. 281,
the lines FR and P^' are parallel to conjugate diameters.
This proposition may also be easily proved geometrically.
For let V and V be the middle points of PJ2 and TR'.
Since V and C are respectively the middle points of EP and J2E',
the hne OF is parallel to FBI, Similarly CV is parallel to FB.
Since GV bisects PE it bisects all chords parallel to PP, i.e. all
chords parallel to GV. So CF' bisects all chords parallel to GV.
Hence CF and GV are in the direction of conjugate diameters and
therefore PP' and JPB,., being parallel to (7F and GV respectively, are
parallel to conjugate diameters.
CONJUGATE DIAMETERS. 261
291. To find the equation to an ellipse referred to a
pair of conjugate diameters.
Let the conjugate diameters be CP and CD (Fig. Art.
286), whose lengths are a' and h' respectively.
If we transform the equation to the ellipse, referred to
its principal axes, to CP and CD as axes of coordinates,
then, since the origin is unaltered, it becomes, by Art. 134,
of the form
Ax' + 2Hxy-\-By'=l (1).
Now the point P, {a, 0), lies on (1), so that
Aa" = \ (2).
So since Q, the point (0, h'), lies on (1), we have
Bh'^ = 1.
Hence A^-jz-, and B = -=-f„.
a^ h^
Also, since CP bisects all chords parallel to CD, there-
fore for each value of x we have two equal and opposite
values of y. This cannot be unless 11=0.
The equation then becomes
^+^-=1
Cor. If the axes be the equiconjugate diameters, the
equation is x^ + y^ = a'^. The equation is thus the same in
form as the equation to a circle. In the case of the ellipse
however the axes are oblique.
292. It will be noted that the equation to the ellipse,
when referred to a pair of conjugate diameters, is of the
same form as it is when referred to its principal axes.
The latter are merely a particular case of a pair of conjugate
diameters.
Just as in Art. 262, it may be shewn that the equation
to the tangent at the point (x, y') is
Similarly for the equation to the polar.
262 COORDINATE GEOMETRY.
Ex. If QVQ' be a double ordinate of the diameter CP, and if the
tangent at Q meet CP in T, then CV . CT=CP'^.
If Q be the point [x', y'), the tangent at it is
Putting y = 0, we have
^rj._a'^_GP^
I.e.
i.e. CV.CT=GP^
EXAMPLES. XXXIV.
1. In the ellipse qF + tt = 1> ^^^ ^^^ equation to the chord which
passes through the point (2, 1) and is bisected at that point.
2. Find, with respect to the elHpse 4:X^ + 7y^=8,
(1) the polar of the point ( - J, 1), and
(2) the pole of the straight line 12a; + 7t/ + 16 = 0.
3. Tangents are drawn from the point {3, 2) to the ellipse
x^ + 4:y'^=9. Find the equation to their chord of contact and the
equation of the straight line joining (3, 2) to the middle point of this
chord of contact.
4. Write down the equation of the pair of tangents drawn to the
ellipse Sx^ + 2y^=5 from the point (1, 2), and prove that the angle
between them is tan"^ — ^ .
a
5. In the ellipse -s + ^=l, write down the equations to the
diameters which are conjugate to the diameters whose equations are
x-y = 0, x + y = 0, y = ^x, and 2/ = -^-
6. Shew that the diameters whose equations are y + Sx = and
iy-x — O, are conjugate diameters of the ellipse 3x^ + 4:y^=5.
7. If the product of the perpendiculars from the foci upon the
polar of P be constant and equal to c^, prove that the locus of P is the
elHpSe 6%2 (^2 + ^2^2^ + g2^4^2^ ^4^4.
8. Shew that the four lines which join the foci to two points P
and Q on an ellipse all touch a circle whose centre is the pole of PQ.
[EXS. XXXIV.] CONJUGATE DIAMETERS. EXAMPLES. 263
9. If the pole of the normal at P lie on the normal at Q, then
shew that the pole of the normal at Q lies on the normal at P.
10. CK is the perpendicular from the centre on the polar of any
point P, and PM is the perpendicular from P on the same polar and
is produced to meet the major axis in L. Shew that (1) CK . PL = b^,
and (2) the product of the perpendiculars from the foci on the polar
= CK.LM.
What do these theorems become when P is on the ellipse ?
11. In the previous question, if PN be the ordinate of P and the
polar meet the axis in T, shew that CL = e^. CN and CT . CN-a^.
12. If tangents TP and TQ be drawn from a point T, whose
coordinates are h and k, prove that the area of the triangle TPQ is
and that the area of the quadrilateral CPTQ is
13. Tangents are drawn to the ellipse from the point
prove that they intercept on the ordinate through the nearer focus a
distance equal to the major axis.
14. Prove that the angle between the tangents that can be drawn
from any point [x^ , y{) to the ellipse is
2ah /■~-±^^yl
tan~i
x{^ + y-^ -a^-b^
15. If T be the point {x^, y-y), shew that the equation to the
straight lines joining it to the foci, S and S\ is
{x-^y - xy-yf - a\^ {y - i/i)^ = 0.
Prove that the bisector of the angle between these lines also
bisects the angle between the tangents TP and TQ that can be drawn
from T, and hence that
lSTP=lS'TQ.
16. If two tangents to an ellipse and one of its foci be given, prove
that the locus of its centre is a straight line.
17. Prove that the straight lines joining the centre to the inter-
sections of the straight line y=mx+ a/ — ^ with the ellipse are
conjugate diameters.
264 COOEDINATE GEOMETRY. [Exs. XXXIV.]
18. Any tangent to an ellipse meets the director circle in -p and d ;
prove that Cp and Gd are in the directions of conjugate diameters of
the ellipse.
19. If CP be conjugate to the normal at Q, prove that GQ is
conjugate to the normal at P.
20. If a fixed straight line parallel to either axis meet a pair of
conjugate diameters in the points K and L, shew that the circle
described on KL as diameter passes through two fixed points on the
other axis.
21. Prove that a chord which joins the ends of a pair of conjugate
diameters of an ellipse always touches a similar ellipse.
22. The eccentric angles of two points P and Q on the ellipse are
01 and 02 ' prove that the area of the parallelogram formed by the
tangents at the ends of the diameters through P and Q is
4a&cosec(0i-02),
and hence that it is least when P and Q are at the end of conjugate
diameters.
23. -A. pair of conjugate diameters is produced to meet the
directrix; shew that the orthocentre of the triangle so formed is at
the focus.
24. If the tangent at any point P meet in the points L and L'
(1) two parallel tangents, or (2) two conjugate diameters,
prove that in each case the rectangle LP . PL' is equal to the square
on the semidiameter which is parallel to the tangent at P.
25. -A. point is such that the perpendicular from the centre on its
polar with respect to the ellipse is constant and equal to c ; shew that
its locus is the elhpse
^2 2/2^1
26. Tangents are drawn from any point on the ellipse -3 + 'rg =1
to the circle x'^ + y'^=r'^ ; prove that the chords of contact are tangents
to the ellipse a^x^^- h^y'^ = r^.
If — = -5 + -s , prove that the lines joining the centre to the points
r^ a^ 0^
of contact with the circle are conjugate diameters of the second
ellipse.
27. GP and CD are conjugate diameters of the ellipse ; prove that
the locus of the orthocentre of the triangle GPD is the curve
2 {hh)'^ + aV)^^ (a2 - 62)2 (^2^2 _ ^2^2)2,
28. If circles be described on two semi-conjugate diameters of the
ellipse as diameters, prove that the locus of their second points of
intersection is the curve 2{x^-\-y'^)'^=a^x^ + h^y'^.
FOUR NORMALS TO AN ELLIPSE. 265
293. To prove that, in general, four normals can he
drawn from any point to an ellipse, and that the sum of the
eccentric angles of their feet is equal to an odd m/ultiple of
two right angles.
The normal at any point, whose eccentric angle is ^, is
r-^ = a^-h^^ de^.
cos <p sm <p
If this normal pass through the point (A, ^), we have
J!L_JL=«v (1).
COS </) sm <jt> ^ '
For a given point (li, k) this equation gives the
eccentric angles of the feet of the normals which pass
through {h, h).
<k
Let tan ^ = ^, so that
cos d> = , = -:, ^ , and sni cb = = .
l + tan^t lH-t-4 '^*
Substituting these values in (1), we have
ah ^ 7, — ok
2/,2
— a^e
\-f 2t
i.e. hkf + 2^3 (ah + aV) + 2i5 (ah - a^e^) -hk = ... (2).
Let ^1, ^2) hi ^^(i h ^6 the roots of this equation, so that,
by Art. 2,
^ah + a?e^ , ,
^X + ^2+^3 + ^4 = -2— ^^- (3),
kk + ^1^3 + ^1^4 + ^2^3 + ^2^4 + kh =^0 (4),
^ ah — a^p} , ,
^2^3^4 + 4^1 + «54^li^2 + ^1^3 = - 2 TT (5),
and ^j^2^o^4=:-l (6),
266 COOEDINATE GEOMETRY.
Hence {Trigonometry, Art. 125), we have
tan {h + ^ + il + h\ - ^1-^3 __ ^1 - ^3 _ ^
•• 2 ='''^+2'
and hence <^i + ^2 + ^3 + </>4 = (2n + 1)??
= an odd multiple of two right angles.
294. We shall conclude the chapter with some ex-
amples of loci connected with the ellipse.
Ex. 1. Find the locus of the intersection of tangents at the ends
of chords of an ellipse, which are of constant length 2c.
Let QR be any such chord, and let the tangents at Q and R meet
in a point P, whose coordinates are {h, k).
Since QR is the polar of P, its equation is
The abscissae of the points in which this straight line meets the
ellipse are given by
\^ a^j ~ h^ V ^y '
L (^ fc^\ _2£i
x^ [W W-\ 2xh , Ti^ ^
If x^ and .^2 be the roots of this equation, i.e. the abscissae of Q
and R, we have
^l + ^2-^-2p:;-^2;i;2' and ^1^2 -j2;,2 + ^2/^2-
If 2/i and 2/2 be the ordinates of Q and R, we have from (1)
a2 "^ 62 -•^'
and 2+^=1'
so that, by subtraction,
_ i'h
2/2 ~ 2/1 — ^2^ V.^2 ~ ^l/'
THE ELLIPSE. EXAMPLES. 267
The condition of the question therefore gives
Hence the point {h, k) always lies on the curve
^ Ka'^y^J-y &2 "^ aP ) V^ + P~-^j'
which is therefore the locus of P.
Ex. 2. Find the locus (1) of the middle points, and (2) of the poles,
of normal chords of the ellipse.
The chord, whose middle point is {h, k), is parallel to the polar of
(h, k) , and is therefore
i^-h)^,+ {y-k)^^=o (1).
If this be a normal, it must be the same as
aa^sec 6 - by cosec d = a^-h^ (2).
We therefore have
gsec 6 _ - 6 cosec 6 a^-b^
"T % ^ /i2 k-^ '
so that cos^:
a^ b^ a^ b^
and sin^=-
h{a^-b^)
fh^ Jf\
\^ '^ by '
y^'^by
k (a2 - fc2)
Hence, by the elimination of d,
fc2\2
The equation to the required locus is therefore
@4:)(^-^i='--^^
e4:r(^3-'-'^>=
Again, if (x^, y-y) be the pole of the normal chord (2), the latter
equation must be equivalent to the equation
^^ + ^^=1 (3).
Comparing (2) and (3), we have
a^sec0_ 6^ cosec ^
^1 vT' '
( a^ W\
so that l = cos2^ + sin2^= — 5+— 5) —
\x^ y^j (a
&2,
1
62)2 •
268 COORDINATE GEOMETRY.
and hence the required locus is
Ex. 3. Chords of the ellipse — 2 + ^2 — ■'■ ^^^'"^^/s touch the concentric
and coaxal ellipse -k + ^ = 1; fend the locus of their poles.
Any tangent to the second ellipse is
yz=mx+ ^Ja^m^ + p^ (1).
Let the tangents at the points where it meets the first ellipse meet
in (h, k). Then (1) must be the same as the polar of {h, k) with
respect to the first ellipse, i.e. it is the same as
a^^b'' ^~" ^''^•
Since (1) and (2) coincide, we have
'h~ k~ Ti
Hence m=--^T, and fja^m^ + jS^ == - .
a^ k
Eliminating m, we have
a4/c2 + P -/^2'
i.e. the point {h, k) lies on the ellipse
^2 1)2
i.e. on a concentric and coaxal ellipse whose semi-axes are — and —
■ a p
respectively.
EXAMPLES. XXXV.
The tangents drawn from a point P to the ellipse make angles 61
and $2 with the major axis ; find the locus of P when
1. ^1 + ^2 i^ constant (=:2a). [Compare Ex. 1, Art. 235.]
2. tan ^i + tan 62 is constant ( = c).
3. tan ^1 - tan d^ is constant ( = d!).
4. tan^ d-i + tan^ ^o is constant ( = X).
[EXS. XXXV.] THE ELLIPSE. EXAMPLES. 269
Find the locus of the intersection of tangents
5. which meet at a given angle a.
6. if the sum of the eccentric angles of their points of contact
be equal to a constant angle 2a.
7. if the difference of these eccentric angles be 120°.
8. if the lines joining the points of contact to the centre be
perpendicular.
9. if the sum of the ordinates of the points of contact be equal to h.
Find the locus of the midSle points of chords of an ellipse
10. whose distance from the centre is the constant length c.
11. which subtend a right angle at the centre.
12. which pass through the given point (/i, Tc).
13. whose length is constant ( = 2c).
14. whose poles are on the auxiliary circle.
15. the tangents at the ends of which intersect at right angles.
16. Prove that the locus of the intersection of normals at the
ends of conjugate diameters is the curve
2 {a?x^ + hhff= {o? - b^ {a^^^ - b^yT-
17. Prove that the locus of the intersection of normals at the ends
of chords, parallel to the tangent at the point whose eccentric angle is
a, is the conic
2 [ax sin a + by cos a) {ax cos a + by sin a) = (a^ - &2)2 g^j^ 2a cos^ 2a.
If the chords be parallel to an equiconjugate diameter, the locus
is a diameter perpendicular to the other equiconjugate.
18. A parallelogram circumscribes the ellipse and two of its
opposite angular points lie on the straight lines x'^ = h^; prove that
the locus of the other two is the conic
^2 y2
(-.:)-■
19. Circles of constant radius c are drawn to pass through the
ends of a variable diameter of the ellipse. Prove that the locus of
their centres is the curve
{x^ + y"^) [a^x^ + b^y^ + a%^) = <?■ {o?x^ + b^y'^) .
20. The polar of a point P with respect to an ellipse touches a
fixed circle, whose centre is on the major axis and which passes
through the centre of the ellipse. Shew that the locus of P is a
parabola, whose latus rectum is a third proportional to the diameter
of the circle and the latus rectum of the ellipse.
21. Prove that the locus of the pole, with respect to the ellipse, of
X
^ v^ 1
any tangent to the auxiliary circle is the curve -4 + n = -^ •
270 COORDINATE GEOMETRY. [EXS. XXXV.]
22. Shew that the locus of the pole, with respect to the auxiliary
circle, of a tangent to the ellipse is a similar concentric ellipse,
whose major axis is at right angles to that of the original ellipse.
23. Chords of the ellipse touch the parabola ay^= -2b^x; prove
that the locus of their poles is the parabola ay^ = 2b^x.
24. Prove that the sum of the angles that the four normals
drawn from any point to an ellipse make with the axis is equal to
the sum of the angles that the two tangents from the same point
make with the axis.
[Use the equation of Art. 268.]
25. Triangles are formed by pairs of tangents drawn from any
point on the ellipse
2 2
a2a;2 + hY = {a^ + ^^f to the ellipse -g + p = 1,
and their chord of contact. Prove that the orthocentre of each such
triangle lies on the ellipse.
26. An ellipse is rotated through a right angle in its own plane
about its centre, which is fixed ; prove that the locus of the point of
intersection of a tangent to the ellipse in its original position with
the tangent at the same point of the curve in its new position is
(a;2 + 2/2) (a;2 + 1,2 _ ^2 _ ^2) ^ 2 {a^ - b^) xy.
27. If y and Z be the feet of the perpendiculars from the foci
upon the tangent at any point P of an ellipse, prove that the tangents
at Y and Z to the auxiliary circle meet on the ordinate of P and that
the locus of their point of intersection is another ellipse.
28. Prove that the directrices of the two parabolas that can be
drawn to have their foci at any given point P of the ellipse and to
pass through its foci meet at an angle which is equal to twice the
eccentric angle of P.
29. Chords at right angles are drawn through any point P of the
ellipse, and the line joining their extremities meets the normal in the
point Q. Prove that Q is the same for all such chords, its
,. , , . a^e2cosa , -a2&e2sina
coordinates being — ^ — 717- and 5 — z^ — .
Prove also that the major axis is the bisector of the angle PGQ,
and that the locus of Q for different positions of P is the ellipse
a;2 y^ _
\^^+by '
CHAPTER XIII.
THE HYPERBOLA.
295. The hyperbola is a Conic Section in which the
eccentricity e is greater than unity.
To find the equation to a hyperbola.
Let ZK be the directrix, 8 the focus, and let SZ be
perpendicular to the directrix.
There will be a point A on AZ^ such that
SA--^e.AZ (ly
272 COORDINATE GEOMETRY.
Since e> 1, there will be another point A\ on /S'^ pro-
duced, such that
SA! = e.A'Z. (2).
Let the length A A! be called 2c&, and let C be the middle
point of AA! .
Subtracting (1) from (2), we have
'la^AA'^e.A:Z-e.AZ
= e\CA' + GZ'\-e\_GA - GZ] = e,2GZ,
i.e. GZ=- , (3).
Adding (1) and (2), we have
e (AZ +A'Z) = SA' + SA = 2GS,
i,e. e.AA' = 2.GS,
and hence GS = ae (4).
Let G be the origin, GSX the axis of x, and a straight
line GT, through G perpendicular to GX, the axis of y.
Let P be any point on the curve, whose coordinates are
X and y, and let FM be the perpendicular upon the directrix,
and PiV the perpendicular on ^^^.
The focus aS' is the point (ae, 0).
The relation JSP"" = e^ . PM^ = e^ . ZN^ then gives
{x — aef + y^ = e^\ x
i.e. x^ — 2aex + a^e^ + 2/^ = &^x^ — ^aex + a^.
Hence x" {e" - I) - y'^ = or {e" - \\
x^ v
i.e. -- -— ^— - ==1 (5).
Since, in the case of the hyperbola, e> 1, the quantity
a^ (e^ — 1) is positive. Let it be called b% so that the equa-
tion (5) becomes
X2 y2
r2-b2=^ (^)'
where b'^=^a'e^-a^^GjS^-GA' (7),
and therefore GS^ = a^ + b^ „ . . . (8).
l2
THE HYPERBOLA. FUNDAMENTAL EQUATION. 273
296. The equation (6) may be written
y^ s^ ^ _x- - a^ _{x — a) {x + a)
PF^_AF.NA'
so that PN^ : AJV . N"!' :: b' : d\
If we put x^O in equation (6), we have y'^- — lf,
shewing that the curve meets the axis CZ in imaginary
points.
Def. The points A and A' are called the vertices of the
hyperbola, C is the centre, -4^' is the transverse axis of the
curve, whilst the line BB' is called the conjugate axis,
where B and B' are two points on the axis of y equidistant
from C, as in the figure of Art. 315, and such that
B'G = CB^h.
297. Since S is the point (ae, 0), the equation referred to the
focus as origin is, by Art. 128,
{x + ae)^ 2/2 _
~"^2 p-1'
i.e. ^'+2--|' + e2-l = 0.
Similarly, the equations, referred to the vertex A and foot of the
directrix Z respectively as origins, wUl be found to be
x^ y^ 2x ^
, a;2 w2 2a; ^ 1
and 1^ +__=!_
a^ b'^ ae e^
^ The equation to the hyperbola, whose focus, directrix, and eccen-
tricity are any given quantities, may be written down as in the case
of the ellipse (Art. 249).
298. There exist a second focus and a second directrix
to the curve.
On SO produced take a point S', such that
SG=CS'^ae,
and another point Z', such that
a
ZC = CZ' =
e
18
274 COORDINATE GEOMETRY.
Draw Z'M' perpendicular to AA^ and let FM be pro-
duced to meet it in M' .
The equation (5) of Art. 295 may be written in the
form
tc^ + ^aex + cC-G^ + 2/^ = ^^ + ^aex + a^,
i.e. {x + aef + 2/^ = e^(-+a3j ,
i.e. S'P^ = e" {Z'G + CNf = e^ . PM\
Hence any point F of the curve is such that its distance
from S' is e times its distance from Z'K\ so that we should
have obtained the same curve if we had started with S' as
focus, Z'K^ as directrix, and the same eccentricity e.
299. The difference of the focal distances of any point
on the hyperbola is equal to the transverse axis.
For (Fig., Art 295) we have
SP = e.PM, and 8'P = e.PM'.
Hence S'P - SP = e{PM' - PM) = e . MM'
= e.ZZ' = 2e.CZ=2a
— the transverse axis AA'.
Also SP = e.PM^e. ZN ^ e.CN-e. CZ^ ex' - a,
and 8'P^e. PM' = e . Z'N =e . C]^+ e .Z'C^bk' + a.,
where x' is the abscissa of the point P referred to the centre
as origin.
300. Latus-rectum of the Hyperbola.
Let LSL' be the latus-rectum, i.e. the double ordinate
of the curve drawn through S.
By the definition of the curve, the semi-latus-rectum SL
= e times the distance of L from the directrix
^e.SZ=e{CS-CZ)
= e . CS — eCZ=ae^ ~a = — ,
a
by equations (3), (4), and (7) of Art. 295.
THE HYPERBOLA. 275
301. To trace the curve
--^=1 (1)
The equation may be written in either of the forms
^=**\/5^ (^)'
or a! = ±fli /^ + 1 (3).
,2
From (2), it follows that, if ^ < cf?^ i.e. if x lie between a
and — (X, then y is impossible. There is therefore no part
of the curve between A and A! .
For all values of oi? > o?- the equation (2) shews that
there are two equal and opposite values of y^ so that the
curve is symmetrical with respect to the axis of x. Also,
as the value of x increases, the corresponding values of y
increase, until, corresponding to an infinite value of x^ we
have an infinite value of y.
For all values of y^ the equation (3) gives two equal
and opposite values to cc, so that the curve is symmetrical
with respect to the axis of y.
If a number of values in succession be given to x^ and
the corresponding values of y be determined, we shall
obtain a series of points, which will all be found to lie on a
curve of the shape given in the figure of Art. 295.
The curve consists of two portions, one of which extends
in an infinite direction towards the positive direction of
the axis of £c, and the other in an infinite direction towards
the neofative end of this axis.
^C5"
302. The quantity — 2 — w — 1 '^^ positive, zero^ or
negative, according as the point {x, y') lies within, u2Jon,
or without, the curve.
Let Q be the point {x, y'), and let the ordinate QW
18—2
276 COORDINATE GEOMETRY.
through Q meet the curve in P, so that, by equation (6) of
Art. 295,
^ _ FIP_ _
and hence —to~ = — ^ — 1-
If Q be within the curve then y, i.e. QN, is less than
Pi\^, sothat |^<_^-, ^.e.<__l.
Hence, in this case, — -j->0, i.e. is positive.
a
t.e
Similarly, if Q be without the curve, then y' > PN, and
we have -^ - ^ — 1 negative.
303. To find the length of any central radius dravjn in
a given direction.
The equation (6) of Art. 295, when tran^rred to polar
coordinates, becomes
(cos- e _Biv? e\
1 COS- 6 sin^e cos-e/b' , . ,\ ,,,
i5=^--^^=-j-.--(;?-ten^^j (1).
This is the equation giving the value of any central
radius of the curve drawn at an inclination 6 to the trans-
verse axis.
52
So long as tan^ ^< -^, the equation (1) gives two equal
and opposite values of r corresponding to any value of 0.
52
For values of tan^ 0> —^, the corresponding values of
a
— are negative, and the corresponding values of r imaginary.
Any radius drawn at a greater inclination than tan~^ -
LENGTH OF A CENTRAL RADIUS. 277
does not therefore meet the curve in any real points, so
that all the curve is included within two straight lines
drawn through C and inclined at an angle ± tan~-^ — to CX.
Writing (1) in the form
7* -—
cos2^(^-tan2(9^
we see that r is least when the denominator is greatest, i.e.
when ^ = 0. The radius vector CA is therefore the least.
Also, when tan ^ = ± - , the value of r is infinite.
a
For values of between and tan"^ - the corresponding
a
positive values of r give the portion ^^ of the curve (Fig.,
Art. 295) and the corresponding negative values give the
portion A'R'.
7
For values of 6 between and — tan~^ - , the positive
a
values of R give the portion AR^, and the negative values
give the portion A'R^.
The ellipse and the hyperbola since they both have a
centre (7, such that all chords of the conic passing through
it are bisected at it, are together called Central Conics.
304. In the hyperbola any ordinate of the curve does
not meet the circle on A A' as diameter in real points.
There is therefore no real eccentric angle as in the case of
the ellipse.
"When it is desirable to express the coordinates of any
point of the curve in terms of one variable, the substitutions
X = a sec ^ and y = b tan <f>
may be used; for these substitutions clearly satisfy the
equation (6) of Art. 295.
The angle ^ can be easily defined geometrically.
On AA! describe the auxiliary circle, (Fig., Art. 306)
278 COORDINATE GEOMETRY.
and from the foot iV of any ordinate NP of the curve draw
a tangent NU to this circle, and join CU. Then
CU=CNco^NCU,
i.e. x = GN=:a^QGNCU.
The angle NCU is therefore the angle <^.
Also HU = CU tan (j> = a ta>ii^,
so that JV^F : JSFU :: b : a.
The ordinate of the hyperbola is therefore in a constant
ratio to the length of the tangent drawn from its foot to
the auxiliary circle.
This angle eft is not so important an angle for the
hyperbola as the eccentric angle is for the ellipse.
305. Since the fundamental equation to the hyper-
bola only differs from that to the ellipse in having — b^
instead of b% it will be found that many propositions for
the hyperbola are derived from those for the ellipse by
changing the sign of b^.
Thus, as in Art. 260, the straight line y — mx + c meets
the hyperbola in points which are real, coincident, or
imaginary, according as
c^> — < a^Tn^ — b^.
As in Art. 262, the equation to the tangent at (x, y'^ is
As in Art. 263, the straight line
y = mx + sJa^TYi^ — b^
is always a tangent.
The straight line
X cos a + y sin a = j^
is a tangent, if p^ = a^ cos^ a — b^ sin^ a.
The straight line Ix + my = n
is a tangent, if ?i^ — aH^ - ¥m^. [Art. 264.]
PROPERTIES OF THE HYPERBOLA.
The normal at the point (x\ y') is, as in Art. 266,
279
306. With some modifications the properties of Arts.
269 and 270 are true for the hyperbola also, if the
corresponding figure be drawn.
In the case of the hyperbola the tangent bisects the
interior, and the normal the exterior, angle between the
focal distances SP and S'P.
It follows that, if an ellipse and a hyperbola have the
same foci aS' and S', they cut at right angles at any common
point P. For the tangents in the two cases are respec-
tively the internal and external bisectors of the angle SPS',
and are therefore at right angles.
307. The equation to the straight lines joining the
points {a sec ^, h tan (j>) and {a sec <^\ h tan ^') can be
shewn to be
X 4>' — ^ y . ct> + c{i 4> + <h'
- cos ^ — 7- sm ^ = cos ^ .
a 2 2 2
280 COORDINATE GEOMETRY.
Hence, by putting (f>' — <^, it follows that the tangent at
the point {a sec ^, h tan ^) is
1- sin </) = cos d).
a 6
It could easily be shewn that the equation to the
normal is
ax sin <^ + by =^{a^ + b^) tan ^.
308. The proposition of Art. 272 is true also for the
hyperbola.
As in Art. 273, the chord of contact of tangents
from (iTj, 2/i) is
^^1 _ y^j _ -1
As in Art. 274, the polar of any point (x^, 2/1) is
^^ y^i __ 1
a^ 52 - •
As in Arts. 279 and 281, the locus of the middle
points of chords, which are parallel to the diameter y = 7nx,
is the diameter y = m-^x, where
b^
a^
The proposition of Art. 278 is true for the hyperbola
also, if we replace b^ by — b\
309. Director circle. The locus of the intersection
of tangents which are at right angles is, as in Art. 271,
found to be the circle x^ + y'^ = a^ — b% i. e. a circle whose
centre is the origin and whose radius is Ja^ — b\
If 5^ < a^, this circle is real.
If b^ — a^, the radius of the circle is zero, and it reduces
to a point circle at the origin. In this case the centre is
the only point from which tangents at right angles can be
drawn to the curve.
If y^ > a^, the radius of the circle is imaginary, so that
there is no such circle, and so no tangents at right angles
can be drawn to the curve.
THE KECTANGULAR HYPERBOLA. 281
310. Equilateral^ or Rectangular^ Hyperbola.
The particular kind of hyperbola in which the lengths
o£ the transverse and conjugate axes are equal is called an
equilateral, or rectangular, hyperbola. The reason for the
name "rectangular" will be seen in Art. 318.
Since, in this case, b = a, the equation to the equilateral
hyperbola, referred to its centre and axes, is x^ -y^ = a\
The eccentricity of the rectangular hyperbola is ^2.
For, by Art. 295, we have, in this case,
, a'' + ¥ 2t*2
a^ a^
so that e — J2.
311. Ex. The perpendiculars from the centre upon the tangent
and normal at any point of the hyperbola -^—j-^ = l meet them in Q
and R. Find the loci of Q and R.
As in Art. 308, the straight line
X cos a + y sin a =p
is a tangent, if p^ — a^ cos^ a-h^ sin^ a.
But p and a are the polar coordinates of Q, the foot of the perpen-
dicular on this straight line from C.
The polar equation to the locus of M is therefore
r2 = a2cos2^-62sin2 6',
i.e., in Cartesian coordinates,
(x^ + y^f^a^x^-h^y^
If the hyperbola be rectangular, we have a = &, and the polar
equation is
r2 = a2 (cos2 d - sin2 6) = a^ cos 26.
Again, by Art. 307, any normal is
axsin(t) + hy = {a^ + h^)ia,n<() (1).
The equation to the perpendicular on it from the origin is
hx-ayB\n = (2).
If we eliminate 0, we shall have the locus of R.
'hx
From (2), we have sind>= — ,
ay'
sin hx
and then tan = / :=r — — j- — .
>/ 1 - sin^ ^Ja^y^— h^x^
Substituting in (1) the locus is
{x^ + 2/2)2 i^aY - b^x^) = (a2 + ^2)2 ^y.
282 COORDINATE GEOMETRY.
EXAMPLES. XXXVI.
Find the equation to the hyperbola, referred to its axes as axes of
coordinates,
1. whose transverse and conjugate axes are respectively 3 and 4,
2. whose conjugate axis is 5 and the distance between whose foci
is 13^
3. whose transverse axis is 7 and which passes through the point
(3, -2),
4. the distance between whose foci is 16 and whose eccentricity
is^/2.
5. In the hyperbola 4a;2- 9i/^ = 36, find the axes, the coordinates
of the foci, the eccentricity, and the latus rectum.
6. Find the equation to the hyperbola of given transverse axis
whose vertex bisects the distance between the centre and the focus.
7. Find the equation to the hyperbola, whose eccentricity is |,
whose focus is {a, 0), and whose directrix is 4:X-dy = a.
Find also the coordinates of the centre and the equation to the
other directrix.
8. Find the points common to the hyperbola 25a;2- 9^/^=225
and the straight line 25a; + 12?/ -45 = 0.
9. Find the equation of the tangent to the hyperbola 4x'^ - 9y^=l
which is parallel to the line 4oy = 5x + l.
10. Prove that a circle can be drawn through the foci of a
hyperbola and the points in which any tangent meets the tangents at
the vertices.
11. An ellipse and a hyperbola have the same principal axes.
Shew that the polar of any point on either curve with respect to the
other touches the first curve.
12. In both an ellipse and a hyperbola, prove that the focal
distance of any point and the perpendicular from the centre upon the
tangent at it meet on a circle whose centre is the focus and whose
radius is the semi-transverse axis.
CC U 5/ '?/ 1
13 Prove that the straight lines — j- = m and - + ^ = - always
^^' ° ah a m
meet on the hyperbola.
14. Find the equation to, and the length of, the common tangent
to the two hyperbolas -^ - p = l and —^- t^=1-
15. In the hyperbola 16^2-9^2 -144^ find the equation to the
diameter which is conjugate to the diameter whose equation is x=2,y.
[EXS. XXXVI.] THE HYPERBOLA. EXAMPLES. 283
16. Find the equation to the chord of the hyperbola
25a;2-16t/2=400
which is bisected at the point (5, 3).
17. In a rectangular hyperbola, prove that
SF.S'P=GF^.
18. the distance of any point from the centre varies inversely as
the perpendicular from the centre upon its polar.
19. if the normal at P meet the axes in G and g, then PG=Pg=PC.
20. *lie angle subtended by any chord at the centre is the
supplement of the angle between the tangents at the ends of the
chord.
21. the angles subtended at its vertices by any chord which is
parallel to its conjugate axis are supplementary.
22. The normal to the hyperbola —, ~ ^ = 1 meets the axes in M
and N, and perpendiculars MP and NP are drawn to the axes ; prove
that the locus of P is the hyperbola
23. If oiie axis of a varying central conic be fixed in magnitude
and position, prove that the locus of the point of contact of a tangent
drawn to it from a fixed point on the other axis is a parabola.
24. If the ordinate MP of a hyperbola be produced to Q, so that
MQ is equal to either of the focal distances of P, prove that the locus
of Q is one or other of a pair of parallel straight lines.
25. Shew that the locus of the centre of a circle which touches
externally two given circles is a hyperbola.
26. On a level plain the crack of the rifle and the thud of the ball
striking the target are heard at the same instant; prove that the
locus of the hearer is a hyperbola.
27. Given the base of a triangle and the ratio of the tangents of
half the base angles, prove that the vertex moves on a hyperbola
whose foci are the extremities of the base.
28. Prove that the locus of the poles of normal chords with
respect to the hyperbola — , - '^ = 1 is the curve
a^ b^
y^a^ - x%^ = (a2 + 62)2 3.2^2,
29. Find the locus of the pole of a chord of the hyperbola which
subtends a right angle at (1) the centre, (2) the vertex, and (3) the
focus of the curve.
30. Shew that the locus of poles with respect to the parabola
y^=^ax of tangents to the hyperbola x^-y^=a^ is the ellipse
4a;2 + 2/2=4a2.
284 COORDINATE GEOMETRY. [EXS. XXXVI.l
31. Prove that the locus of the pole with respect to the hyperbola
—„ = \ of any tangent to the circle, whose diameter is the line
a?- W-
♦
3j 11 1
ng the foci, is the ellipse — 4 + ri =
a^ ¥ a^+b^
32. Prove that the locus of the intersection of tangents to a
hyperbola, which meet at a constant angle /3, is the curve
33. From points on the circle x^ + y^=a^ tangents are drawn to
the hyperbola x^ - y'^=a^; prove that -the locus of the middle points of
the chords of contact is the curve
(a;2 - 1/2)2 _ ^2 ^^2 ^ y2j_
34. Chords of a hyperbola are drawn, all passing through the
fixed point {h, 1c) ; prove that the locus of their middle points is a
hyperbola whose centre is the point ( ^ , - J , and which is similar to
either the hyperbola or its conjugate.
312. Asymptote. Def. An asymptote is a straight
line, which meets the conic in two points both of which are
situated at an infinite distance, but which is itself not alto-
gether at infinity.
313. To find the asymptotes of the hyperbola
As in Art. 260, the straight line
y = 7nx + c (1)
meets the hyperbola in points, whose abscissae are given by
the equation
x" {¥ - a'Tiv')- 2a^mcx - a^ (c^ + 6^) = Q (2).
If the straight line (1) be an asymptote, both roots of (2)
must be infinite.
Hence (C. Smith's Algebra, Art. 123), the coefiicients of
x^ and X in it must both be zero.
We therefcwre have
h^ — a^jn^ = 0, and a^mc — 0.
ASYMPTOTES. 285
Hence m = =*=-, and c = 0.
a
Substituting these values in (1), we have, as the^fc
quired equation, ^^
h
-W =: =fc - aj.
^ a
There are therefore two asymptotes both passing
through the centre and equally inclined to the axis of x,
the inclination being
tan"-^ — .
a
The equation to the asymptotes, written as one equa-
tion, is
Cor. For all values of c one root of equation (2) is
infinite if 7?2 = ± — . Hence any straight line, which is
parallel to an asymptote, meets the curve in one point at
infinity and in one finite point.
314. That the asymptote passes through two coincident points
at infinity, i. e. touches the curve at infinity, may be seen by finding
the equations to the tangents to the curve which pass through any
point f rcj , — ajj^ j on the asymptote y=-x.
As in Art. 305 the equation to either tangent through this point is
y = mx + Ja^m^ - b^^
where - ar, = mx, -{• Ja^m^ - 6^
a ^ ^
i.e. on -clearing of surds,
m2 {x^^ - a2) - 2m - x-,^+ (x.^ + a^) -,=0.
One root of this equation is m = -, so that one tangent through
the given point \^y=:- x, i.e. the asymptote itself.
286
COORDINATE GEOMETRY.
315. Geometrical construction fo7' the asymptotes.
Let A' A be the transverse axis, and along the conju-
gate axis measure off CB and CB\ each equal to h.
Through B and B' draw parallels to the transverse axis
and through A and A' parallels to the conjugate axis, and
let these meet respectively in K^, K^^ K^, and K^, as in the
figure.
Clearly the equations of K^CK^ and K^CK^ are
h , b
y = — x, and y = x,
^ a ' ^ a '
and these are therefore the equations of the asymptotes.
316. Let any double ordinate FlSfP' of the hyperbola
be produced both ways to meet the asymptotes in Q and Q\
and let the abscissa CN be x.
Since P lies on the curve, we have, by Art. 302,
NP^-slx'^-a''
a
ASYMPTOTES. 287
Since Q is on the asymptote whose equation is 3/ = - x,
we have NQ = - £c
a
Hence PQ = NQ - NP = ^- (x' - slx'^ - a\
and QP' ^^- (x' + sj^-a\
a ^ '
Therefore PQ .QP' = ^ jcc'^ - (x'^ - a')} = h\
Hence, if from any point on an asymptote a straight
line be drawn perpendicular to the transverse axis, the
product of the segments of this line, intercepted between
the point and the curve, is always equal to the square on
the semi-conjugate axis.
Again,
PQj-{^-^W^r^) = ^
« ax' + Jx'''-a''
ah
x' + \/x^ — a^
PQ is therefore always positive, and therefore the
part of the curve, for which the coordinates are positive,
•is altogether between the asymptote and the transverse
axis.
Also as X increases, i. e. as the point P is taken further
and further from the centre C, it is clear that PQ con-
tinually decreases ; finally, when x' is infinitely great, PQ
is infinitely small.
The curve therefore continually approaches the asymp-
tote but never actually reaches it, although, at a very great
distance, the curve would not be distinguishable from the
asymptote.
This property is sometimes taken as the definition of an
asymptote.
317. If SF be the perpendicular from ^S' upon an
asymptote, the point F lies on the auxiliary circle. This
288 COORDINATE GEOMETRY,
follows from the fact that the asymptote is a tangent,
whose point of contact happens to lie at infinity, or it may
be proved directly.
For
GF= CS cos FCS =CS.%= j¥+¥ . -.-^^ - a.
Also Z being the foot of the directrix, we have
GA^= CS.cz, (Art. 295)
and hence CF^ = CS . CZ, i.e. CS : CF :: CF : CZ.
By Euc. YI. 6, it follows that l CZF= l CFS= a right
angle, and hence that F lies on the directrix.
Hence the perpendiculars from the foci on either asymptote
meet it in the sam,e points as the corresponding directrix,
and the common points of intersection lie on the auxiliary
circle.
318. Equilateral or Rectangular Hyperbola.
In this curve (Art. 310) the quantities a and b are equal.
The equations to the asymptotes are therefore y=.^x, i.e.
they are inclined at angles =t 45° to the axis of x, and hence
they are at right angles. Hence the hyperbola is generally
called a rectangular hyperbola.
319. Conjugate Hyperbola. The hyperbola which
has BB' as its transverse axis, and A A' as its conjugate
axis, is said to be the conjugate hyperbola of the hyperbola
whose transverse and conjugate axes are respectively AA'
and BB'.
Thus the hyperbola
¥ a'~^ ^ ^'
is conjugate to the hyperbola
^-^ = 1 (2).
a' b' ^ ^
Just as in Art. 313, the equation to the asymptotes of
F~a'
11^ ex?
X-
THE HYPERBOLA. CONJUGATE DIAMETERS. 289
which, by the same article, is the equation to the asymp-
totes of (2).
Thus a hyperbola and its conjugate have the sam.e
asymptotes.
The conjugate hyperbola is the dotted curve in the
figure of Art. 323.
320. Intersections of a hyperbola with a pair of con-
jugate diameters.
The straight line y = m^x intersects the hyperbola
a^ b'
in points whose abscissae are given by
i.e. by the equation a^ — — — - .
^ 0^ — a^m^
The points are therefore real or imaginary, according as
a^TYi^ is < or > 6^,
i.e. according as
m^ is numerically < or > - (1),
a
i.e. according as the inclination of the straight line to the
axis of X is less or greater than the inclination of the
asymptotes.
Now, by Art. 308, the straight lines y - tyi^x and y = m^x
are conjugate diameters if
^' /ox
»^1^2 = ^ (2).
Hence one of the quantities m^ and m^ must be less
than - and the other greater than - .
a a
Let mj be < -, so that, by (1), the straight line y = m^x
meets the hyperbola in real points.
L. 19
290 COOEDINATE GEOMETRY.
Then, by (2), m^ must be > - , so that, by (1), the straight
CL
line y = m^x will meet the hyperbola in imaginary points.
It follows therefore that only one of a pair of conjugate
diameters meets a hyperbola in real points.
321. If a pair of diameters he conjugate with respect
to a hyperbola, they will he conjugate with respect to its con-
jugate hyperhola.
Eor the straight lines y = tyi^ and y = m^ are conjugate
with respect to the hyperbola
a^ y"
a
-|.-1 (1).
if ^i'^2==-2 (2).
Cv
Now the equation to the conjugate hyperbola only
differs from (1) in having — a^ instead of a^ and - h^ instead
of h% so that the above pair of straight lines will be con-
jugate with respect to it, if
'^™= = l-^ = a^ (^)-
But the relation (3) is the same as (2).
Hence the proposition.
322. If a pair of diameters he conjugate with respect
to a hyperhola, one of them meets the hyperhola in real points
and the other meets the conjugate hyperhola in real points.
Let the diameters be 3/ = m^x and y = m^x, so that
h^
m^m^ = — .
a^
As in Art. 320 let mi < - , and hence m- > - , so that the
^ a' ^ «'
straight line y — m^x meets the hyperbola in real points.
Also the straight line y — m^x meets the conjugate
hyperbola ^ -^-\ in points whose abscissae are given by
a
a^y
the equation x^ ( -zj A — \y i.e. by .t^ =
ra^ a^ — h"^'
THE HYPERBOLA. CONJUGATE DIAMETERS. 291
Since mo > - , these abscissae are real.
" a
Hence the proposition.
323. If aiKiir of conjugate diameters meet the hyperbola
and its conjugate in P and D, then (1) CP^ — CD^ = a^ — 6^,
and (2) the tangents at P, D and the other ends of the
diameters passing through them form a 'parallelogram whose
vertices lie on the asymptotes and whose area is constant.
Let P be any point on the hyperbola — — ^ = 1 whose
coordinates are (a sec cf), h tan ^).
The equation to the diameter CP is therefore
h tan 4> ^ • I
y = J x = x . - sm <p.
a sec <^
a
By Art. 308, the equation to the straight line, Avhich
is conjugate to CP, is
b
y ^x — . — -
a sm <fi
19—2
292 COORDINATE GEOMETRY.
This straight line meets the conjugate hyperbola
in the points (a tan <^, h sec ^), and (— a tan (j>, —h sec ^) so
that D is the point {a tan <^, 6 sec ^).
We therefore have
CP" - a' sec^ «^ + 6Han2 </,,
and CZ)^ - a^ tan^ <^ + ¥ sec^ <^.
Hence
CP2 _CD^^{a''- h^) (sec2 ^ - tan^ </>) = c*2 - 51
Again, the tangents at P and Z> to the hyperbola and
the conjugate hyperbola are easily seen to be
f sind) =^cos d), (1).
a ^ \ /'
If Q^
and 7 sind) = cosd) (2).
a ^ '
These meet at the point
X y cos ^
a h 1 — sin «^ *
This point lies on the asymptote CL.
Similarly, the intersection of the tangents at P and D '
lies on CL-[^ that of tangents at D' and F' on GL' ^ and
those at D and F' on CZj.
If tangents be therefore drawn at the points where a
pair of conjugate diameters meet a hyperbola and its
conjugate, they form a parallelogram whose angular points
are on the asymptotes.
Again, the perpendicular from C on the straight line (1)
cos ^ ah cos
y
1 1 . , , V*" + a' sin' A
a6 «6 ah
sJhHec^<j> + d'tQ.n^4> CD FK'
THE CONJUGATE HYPERBOLA. 293
SO that PK X perpendicular from G on TK ~ ah,
i.e. area of the parallelogram CPKD — ah.
Also the areas of the parallelograms CPKD, CDK^P\
CP'K'D', and CB'K;P are all equal.
The area KK^K'K^ therefore = 4a6.
Cor. PK^ CD = D'C = K^P, so that the portion of a
tangent to a hyperbola intercepted between the asymptotes
is bisected at the point of contact.
324. Relation hefweert the equation to the hyperhola,
the equation to its asymptotes, and the equation to the conju-
gate hyperbola.
The equations to the hyperbola, the asymptotes, and the
conjugate hyperbola are respectively
5-F = i • W-
%-%-' -(2)-
and-^-|^ = -l (3).
a^ 0"
We notice that the equation (2) differs from equation (1)
by a constant, and that the equation (3) differs from (2) by
exactly the same quantity that (2) differs from (1).
If now we transform the equations in any way we
please — by changing the origin and directions of the axes —
by the most general substitutions of Art. 132 and by
multiplying the equations by any — the same — constant,
we shall alter the left-hand members of (1), (2), and (3) in
exactly the same way, and the right-hand constants in the
equations will still be constants, and differ in the same way
as before.
Hence, whatever be the form of the equation to a
hyperbola, the equation to the asymptotes only differs from
it by a constant, and the equation to the conjugate
hyperbola differs from that to the asymptotes by the same
constant.
294 COORDINATE GEOMETRY.
325. As an example of the foregoing article, let it be required
to find the asymptotes of the hyperbola
Sx^-5xy-2y^ + 5x+ny-8 = ...(1).
Since the equation to the asymptotes only differs from it by a
constant, it must be of the form
Sx^-5xy-2y^ + 5x + lly + c = (2).
Since (2) represents the asymptotes it must represent two straight
lines. The condition for this is (Art. 116)
3(-2)c + 2.|.-V-(-f)-3(W-(-2)(t)2-c(-t)2=0,
i.e. c=-12.
The equation to the asymptotes is therefore
3a;2 - 5xy - 2tf + 5a; + 11?/ - 12 = 0,
and the equation to the conjugate hyperbola is
3a;2 - 5xy -2i/ + 5x + lly -16 = 0.
326. As another example we see that the equation to any
hyperbola whose asymptotes are the straight lines
Ax + By + G = and A-^x + B-^y + 0-^ = 0,
is {Ax + By + C){A^x + B^y + Gj) = \'' (1),
where X is any constant.
For (1) only differs by a constant from the equation to the
asymptotes, which is
{Ax + By + C){A^x + B^y + G^) = (2).
If in (1) we substitute - A^ for X^ ^e shall have the equation to its
conjugate hyperbola.
It follows that any equation of the form
[Ax + By + C) {A^x + B-^y + Oj) = X^
represents a hyperbola whose asymptotes are
Ax + By + C = 0, and A^x + B^y + G^ = 0.
Thus the equation x{x + y) = a^ represents a hyperbola whose
asymptotes are x = and x + y = 0.
Again, the equation x'^ + 2xy cot 2a - y^ = a'^,
i.e. {x cot a- y) {x ta.n a + y) = a^,
represents a hyperbola whose asymptotes are
X cot a-y = 0, and x tan a + y = 0.
327. It would follow from the preceding articles that the
equation to any hyperbola whose asymptotes are x = and ?/ = is
a;2/ = const.
THE HYPERBOLA. EXAMPLES. 295
The constant could be easily determined in terms of the semi-
transverse and semi-conjugate axes.
In Art. 328 we shall obtain this equation by direct transformation
from the equation referred to the principal axes.
EXAMPLES. XXXVII.
1. Through the positive vertex of the hyperbola a tangent is
drawn; where does it meet the conjugate hyperbola?
2. lie and e' be the eccentricities of a hyperbola and its conjugate,
prove that _ + _=!.
3. Prove that chords of a hyperbola, which touch the conjugate
hyperbola, are bisected at the point of contact.
4. Shew that the chord, which joins the points in which a pair of
conjugate diameters meets the hyperbola and its conjugate, is parallel
to one asymptote and is bisected by the other.
5. Tangents are drawn to a hyperbola from any point on one of
the branches of the conjugate hyperbola; shew that their chord of
contact will touch the other branch of the conjugate hyperbola.
6. A straight line is drawn parallel to the conjugate axis of a
hyperbola to meet it and the conjugate hyperbola in the points P and
Q ; shew that the tangents at P and Q meet on the curve
6^'
4x^
and that the normals meet on the
axis
of a;.
7. From a point G on the transverse axis GL is drawn perpen-
dicular to the asymptote, and GP a normal to the curve at P. Prove
that LP is parallel to the conjugate axis.
8. Find the asymptotes of the curve 2x'^ + 5xy + 2ij^ -\-4:X + 5y = 0,
and find the general equation of all hyperbolas having the same
asymptotes.
9. Find the equation to the hyperbola, whose asymptotes are the
straight lines x + 2y + 3 = 0, and Sx + 4y + 5 = 0, and which passes
through the point (1,-1).
Write down also the equation to the conjugate hyperbola.
10. In a rectangular hyperbola, prove that CP and CD are equal,
and are inclined to the axis at angles which are complementary.
296
COORDINATE GEOMETRY. [ExS. XXXVII.]
11. C is the centre of the hyperbola -3 - t2 = 1 and the tangent at
any point P meets the asymptotes in the points Q and R. Prove that
the equation to the locus of the centre of the circle circumscribing
the triangle CQB is 4 {a^x^ - b^) = (a^ + b^f.
12. A series of hyperbolas is drawn having a common transverse
axis of length 2a. Prove that the locus of a point P on each hyper-
bola, such that its distance from the transverse axis is equal to its
distance from an asymptote, is the curve {x^-y^)'^=4:X^{x^-a^).
328. To jind the equation to a hyperbola referred to its
asymptotes.
Let P be any point on the hyperbola, whose equation
referred to its axes is
,2 '" ^
.(1).
a" 52
Draw PH parallel to one asymptote CL to meet the
other CK' in ZT, and let CH and HP be h and k respec-
tively. Then h and k are the coordinates of P referred to
the asymptotes.
Let a be the semi-angle between the asymptotes, so that,
by Art. 313, tan a = — ,
sm a cos a
1
and hence , ,
Draw I{]V perpendicular to the transverse axis, and IIP
parallel to the transverse axis, to meet the ordinate PM of
the point P in P.
ASYMPTOTES AS AXES. 297
Then, since PH and HR are parallel respectively to CL
and CM, we have l PHR ^LLCM=a.
Hence GM^ GN+HR=CHcob a + HPco^a
and MP^BP- UN ^ HP sin a - C^sin a
Therefore, since Cilf and MP satisfy the equation (1),
- "we have
Hence, since (A, ^) is any point on the hyperbola, the
required equation is
xy = — 4— ■
This is often written in the form xy — c^, where 4c^
equals the sum of the squares of the semiaxes of the
hyperbola.
Similarly, the equation to the conjugate hyperbola is,
when referred to the asymptotes,
^y = J—.
329. To find the equation to the tangent at any point
of the hyperbola xy = c^.
Let (cc', y') be any point P on the hyperbola, and
{x", y") a point Q on it, so that we have
^y = c2 (1),
and x"y" = 0^ (2).
The equation to the line PQ is then
2/ -y'=fc|^ (»-=«') (3).
298 COORDINATE GEOMETRY.
But, by (1) and (2), we have
c'
y"-v' »"
x'
& x'-x"
c^
rJi __ y ~ y
-x'
JU JU tAj ~~~ JU
" x'^'
Hence the equation (3) becomes
c
tAJ %AJ
Let now the point Q be taken indefinitely near to P, so
that x' — X ultimately, and therefore, by Art. 149, FQ
becomes the tangent at P.
Then (4) becomes
y
The required equation is therefore
x'i/ + xy — Ixy — 2g^ (5).
The equation (5) may also be written in the form
% + K-^ -^ (6).
X y
330. The tangent at any point of a hyperbola cuts off a triangle
of constant area from the asymptotes, and the portion of it intercepted
hetioeen the asymptotes is bisected at the point of contact.
Take the asymptotes as axes and let the equation to the hyperbola
be xy = c^.
The tangent at any point P is — + — , = 2.
X y
This meets the axes in the points [2x' , 0) and (0, 2?/').
If these points be L and U, and the centre be C, we have
GL = 'ix', 2in(iGL' = 2y'.
If 2a be the angle between the asymptotes, the area of the triangle
LGL' = ^CL . OL' sin 2a = 2a;'2/'sin 2a= — ^— . 2 sinacos a = a6.
(Art. 328.)
Also, since L is the point (2a;', 0) and L' is (0, 2y'), the middle
point of LL' is (x', y'), i.e. the point of contact P.
ASYMPTOTES AS AXES. 299
331. As in Art. 274, the polar of any point (x^, y^
with respect to the curve can be shewn to be
Since, in general, the point (aj^, y^ does not lie on the
curve the equation to the polar cannot be put into the form
(6) of Art. 329.
332. The equation to the normal at the point {x\ y')
is y-y' =^m{x— x'), where m is chosen so that this line is
perpendicular to the tangent
y 2c
y = --,x + —r'
^ X X
If 0) be the angle between the asymptotes we then
obtain, by Art. 93,
x' — y cos 0)
m^— -, ,
y — X cos 0)
so that the required equation to the normal is
y (if' — X cos w) — X (x — y cos w) = y"^ — x'.
Also cos w = cos 2a = cos a — sm a =r — - — ~
L 0^ + 6 .
If the hyperbola be rectangular, then co = 90°, and the
equation to the normal becomes xx — yy' — x'^ — y'^.
333. Equation referred to the asymptotes.
One Variable.
The equation xy = c^ is clearly satisfied by the substitu-
tion x = ct and y = - .
Hence, for all values of t, the point whose coordinates
are ict, - j lies on the curve, and it may be called the point
The tangent at the point "^" is by Art. 329,
X
300 COORDINATE GEOMETRY
Also the normal is, by the last article,
c
1
c
y (l — f cos iji)—x {f — cos (o) == - (1 - ^*),
or, when the hyperbola is rectangular,
The equations to the tangents at the points ^'t" and " t"
are
-+yt^=. 2c, and - + yt^= 2c,
h h
and hence the tangents meet at the point
Vi + ^2 ' ^1 + tj '
The line joining " t^" and " ^2/' which is the polar of this
point, is therefore, by Art. 331,
X + yt^t^ = c(ti + 1^.
This form also follows by writing down the equation
to the straight line joining the points
(ct,, g and (ct,, Q
334. Ex. 1. If a rectangular hyperbola circumscribe a triangle,
it also passes through the orthocentre of the triangle.
Let the equation to the curve referred to its asymptotes be
a:y = c'^ (1).
Let the angular points of the triangle be P, Q, and R, and let their
coordinates be
j-espectively.
As in the last article, the equation to QR is
x + yt2ts=c{t^ + ts).
The equation to the straight line, through P perpendicular to QR^
is therefore
h
I.e.
y + ctihh=hh[f + jjj] (2)-
ASYMPTOTES AS AXES. 301
Similarly, the equation to the straight line through Q perpendicular
to RP is
2/ + c«iU3=¥ir^ + j-7T] (^)-
The common point of (2) and (3) is clearly
{~4j^ "'''''') ^'^'
and this is therefore the orthocentre.
But the coordinates (4) satisfy (1). Hence the proposition.
Also if ( ct^,-\ he the orthocentre of the points " ij," " t^" and
" ig," we have t^t^t^t^— - 1.
Ex. 2. If a circle and the rectangular hyperbola (xy = c^ meet in
the four points "ti," "f2>" "^3>" ^nd ^^t^^" prove that
(1) ti«2^3«4=l,
(2) the centre of mean position of the four points bisects the
distance between the centres of the two curves,
and (3) the centre of the circle through the points "ij," "f2»" "^g" ^^
Let the equation to the circle be
x^ + y^-2gx-^y + k = 0,
so that its centre is the point {g,f).
Any point on the hyperbola is (ct,jj. If this lie on the circle,
we have cH^ + -^-2gct-2f- + k = 0,
so that i4 - 2^ t^+-t^--^t +1 = (1).
c c^ c ^ '
If t^ , fg, *3, and ^4 be the roots of this equation, we have, by Art. 2,
hhhh='^ (2),
h + h + t, + t,=^ (3),
2f
and *2M4 + *3*4*1 + *4M2+M2*3= (4).
Dividing (4) by (2), we have
11112/
302 COORDINATE GEOMETRY.
The centre of the mean position of the four points,
i. e. the point || {t^ + *2 + «3 + «4). | (^- + ^ + ^ + l^f ,
is therefore the point ( k j ^ ) ? ^^^ tliis is the middle point of the line
joining (0, 0) and (^,/)
1
Also, since i^ = -^-^-^ , we have
i-icgi's
= l{t, + h + h+^), and /=|(l + l + i + MA).
Again, since ^1*2*3*4 = 15 "^^ have product of the abscissae of the
four points = product of their ordinates = c^.
EXAMPLES. XXXVIII.
a^ + b^
1. Prove that the foci of the hyperbola xy= — j — - are given by
2. Shew that two concentric rectangular hyperbolas, whose axes
meet at an angle of 45°, cut orthogonally.
3. A straight line always passes through a fixed point; prove
that the locus of the middle point of the portion of it, which is
intercepted between two given straight lines, is a hyperbola whose
asymptotes are parallel to the given lines.
4. If the ordinate NP at any point P of an ellipse be produced to
Q, so that NQ is equal to the subtangent at P, prove that the locus of
^ is a hyperbola.
5. From a point P perpendiculars PM and PN are drawn to two
straight lines OM and ON. If the area OMPN be constant, prove
that the locus of P is a hyperbola.
6. A variable line has its ends on two lines given in position awd
passes through a given point ; prove that the locus of a point which
divides it in any given ratio is a hyperbola.
7. The coordinates of a point are a tan (^ + a) and 6tan(^ + j8),
where 6 is variable ; prove that the locus of the point is a hyperbola.
8. A series of circles touch a given straight line at a given point.
Prove that the locus of the pole of a given straight line with regard to
these circles is a hyperbola whose asymptotes are respectively a
parallel to the first given straight line and a perpendicular to the
second.
[EXS. XXXVIII.] THE HYPEKBOLA. EXAMPLES. 303
9. If a right-angled triangle be inscribed in a rectangular hyper-
bola, prove that the tangent at the right angle is the perpendicular
upon the hypothenuse.
10. In a rectangular hyperbola, prove that all straight lines, which
subtend a right angle at a point P on the curve, are parallel to the
normal at P.
11. Chords of a rectangular hyperbola are at right angles, and
they subtend a right angle at a fixed point ; prove that they inter-
sect on the polar of 0.
12. Prove that any chord of a rectangular hyperbola subtends
angles which are equal or supplementary (1) at the ends of a perpen-
dicular chord, and (2) at the ends of any diameter.
13. In a rectangular hyperbola, shew that the angle between a
chord PQ and the tangent at P is equal to the angle which PQ
subtends at the other end of the diameter through P.
14. Show that the normal to the rectangular hyperbola xy — c- at
the point "t" meets the curve again at a point "t"' such that
i¥=-l.
15. If Pj, P25 ^^cl P3 be three points on the rectangular hyperbola
xy = c^, whose abscissae are x-,^, x^ , and x^ , prove that the area of the
triangle P^P^P^ is
^ «^-i J;o wo
and that the tangents at these points form a triangle whose area is
2^2 (^2~^3) (^3~^l) (^l~^2)
(^2 + %) (^3 + ^1) {^1 + ^2) "
16. Find the coordinates of the points of contact of common
tangents to the two hyperbolas
x^-y^=Sa^ and xy = 2a^.
17. The transverse axis of a rectangular hyperbola is 2c and the
asymptotes are the axes of coordinates ; shew that the equation of the
chord which is bisected at the point (2c, 3c) is Sx + 2y = 12c.
18. Prove that the portions of any line which are intercepted
between the asymptotes and the curve are equal.
19. Shew that the straight lines drawn from a variable point on
the curve to any two fixed points on it intercept a constant distance on
either asymptote.
20. Shew that the equation to the director circle of the conic
xy = c^is x^ + 2xy cos (>)+y^ = 4ie^ cos (a.
21. Prove that the asymptotes of the hyperbola xy = hx + 1cy are
x = k and y = h.
304 COORDINATE GEOMETRY. [EXS.
22. Shew that the straight line y = mx + c\/ - m always touches the
hyperbola xy = c'^, and that its point of contact is I /-— - , of - 711
23. Prove that the locus of the foot of the perpendicular let fall
from' the centre upon chords of the rectangular hyperbola xy = c^
which subtend half a right angle at the origin is the curve
24. A tangent to the parabola x^ = 4a?/ meets the hyperbola xy = k^
in two points P and Q. Prove that the middle point of PQ hes on a
parabola.
25. If a hyperbola be rectangular, and its equation be xy = c^,
prove that the locus of the middle points of chords of constant length
M is {x'^ + 2/^) {xy - c^) = d^xy.
26. Shew that the pole of any tangent to the rectangular hyper-
bola xy^c^, with respect to the circle x^ + y^ = a^, lies on a concentric
and similarly placed rectangular hyperbola.
27. Prove that the locus of the poles of all normal chords of the
rectangular hyperbola xy = c^ is the curve
{a;2- 2/2)2 + 4c2^i/ = 0.
28. Any tangent to the rectangular hyperbola 4:xy = ah meets the
o o
ellipse — + — „= 1 in the points P and Q ; prove that the normals at P
^ a?" h^
and Q to the ellipse meet on a fixed diameter of the ellipse.
29. Prove that triangles can be inscribed in the hyperbola xy = c^,
whose sides touch the parabola y^=4iax.
30. A. point moves on the given straight line y=mx; prove that
the locus of the foot of the perpendicular let fall from the centre upon
its polar with respect to the ellipse -2 + ^ = '^ is a rectangular
hyperbola, one of whose asymptotes is the diameter of the ellipse
which is conjugate to the given straight line.
31. A quadrilateral circumscribes a hyperbola; prove that the
straight line joining the middle points of its diagonals passes through
the centre of the curve.
32. A, B, C, and D are the points of intersection of a circle and a
rectangular hyperbola. If AB pass through the centre of the hyper-
bola, prove that CD passes through the centre of the circle.
33. If a circle and a rectangular hyperbola meet in four points P,
Q, r] and S, shew that the orthocentres of the triangles QRS, RSP^
SPQ, and PQR^iiso lie on a circle.
Prove also that the tangents to the hyperbola at R and S meet
in a point which lies on the diameter of the hyperbola which is at
right angles to PQ.
XXXVIII.] THE HYPERBOLA. EXAMPLES. 305
34. A series of hyperbolas is drawn, having for asymptotes the
principal axes of an ellipse; shew that the common chords of the
hyperbolas and the ellipse are all parallel to one of the conjugate
diameters of the ellipse.
35. A circle, passing through the centre of a rectangular hyperbola,
cuts the curve in the points A, B, G, and D ; prove that the circum-
circle of the triangle formed by the tangents at A, B, and C goes
through the centre of the hyperbola, and has its centre at the point
of the hyperbola which is diametrically opposite to D.
36. Given five points on a circle of radius a; prove that the
centres of the rectangular hyperbolas, each passing through four of
these points, all lie on a circle of radius - .
37. If a rectangular hyperbola circumscribe a triangle, shew that
it meets the circle circumscribing the triangle in a fourth point, which
is at the other end of the diameter of the hyperbola which passes
through the orthocentre of the triangle.
Hence prove that the locus of the centre of a rectangular hyper-
bola which circumscribes a triangle is the nine-point circle of the
triangle.
38. Two rectangular hyperbolas are such that the asymptotes of
one are parallel to the axes of the other and the centre of each lies on
the other. If any circle through the centre of one cut the other again
in the points P, Q, and jR, prove that PQR is a triangle such that each
side is the polar of the opposite vertex with respect to the first
hyperbola.
20
CHAPTER XIV.
POLAR EQUATION OF A CONIC SECTION, ITS FOCUS
BEING THE POLE.
335. Let S be the focus, A the vertex, and ZM the
directrix ; draw SZ perpendicular to ZM.
Let ZS be chosen as the positive direction of the
initial line, and produce it to X,
Take any point P on the
curve, and let its polar co-
ordinates be r and 6, so that
we have
SP = r, and iXSP^O.
Draw PN perpendicular
to the initial line, and PM
perpendicular to the directrix.
Let SL be the semi-latus-
rectum, and let SL = I.
Since SL = e , SZ^ we have
szJ-,
e
Hence
r = SP = e.PM=e,ZN
= e{ZS + SN)
I
= ei- + SP. cosO
Therefore
r =
= l + e.r. cos 0.
1
1 — 6 008 6^
.(1).
THE POLAR EQUATION, FOCUS BEING POLE. 307
This, being the relation holding between the polar
coordinates of any point on the curve, is, by Art. 42, the
required polar equation.
Cor. If SZ be taken as the positive direction of the initial line and
the vectorial angle measured clockwise, the equation to the curve is
I
r= .
1 + e cos d
336. If the conic be a parabola, we have e = l, and the equation
I I I ^e
IS r = z = — ^— = - cosec^ - .
1-cos^ ^ . ^d 2 2
2 sm^ -
If the initial line, instead of being the axis, be such that the axis
is inclined at an angle y to it, then, in the previous article, instead of
6 we must substitute d -y.
The equation in this case is then
-=l-ecos(^-7).
337. To trace the curve — = \—e cos 6.
r
Case I. e = 1^ so that the equation is - = 1 - cos 6.
When is zero, we have - — 0, so that r is infinite. As
r
increases from 0° to 90°, cos^ decreases from 1 to 0,
and hence - increases from to 1, i.e. r decreases from
r
infinity to I.
As increases from 90° to 180°, cos^ decreases from
to — 1, and hence - increases from 1 to 2, i.e. r decreases
r
from I to ^l.
Similarly, as changes from 180° to 270°, r increases
from - to I, and, as changes from 270° to 360°, r increases
from ^ to CO .
The curve is thus the parabola co FPLAL'P'F' oo of
Art. 197.
20—2
308 COORDINATE GEOMETRY.
Case II. e<l. When is zero, we have - = 1— e,
r
i.e. r— r . This gives the point A' in the figure of Art.
247.
As 6 increases from 0° to 90°, cos B decreases from 1 to
0, and therefore 1— ecos^ increases from 1 -e to 1, i.e. -
r
increases from 1— e to 1, i.e. r decreases from to I.
1— e
"We thus obtain the portion A'PBL.
As 6 increases from 90° to 180°, cos^ decreases from
to — 1, and therefore 1—e cos increases from 1 to 1 + e,
i.e. - increases from 1 to 1 + e, i.e.r decreases from I to r .
r 1 +e
We thus obtain the portion LA of the curve, where
SA = ^.
Similarly, as increases from 180° to 270° and then to
360°, we have the portions AL' and L'B'P'A'.
Since cos 6 = cos (— 6) = cos (360° — 6), the curve is sym-
metrical about the line SA'.
Case III. e > 1. When is zero, 1 -e cos 6 is equal
to 1—e, i.e. — (e— 1), and is therefore a negative quantity,
since e > 1. This zero value of gives r = — I — (e — 1).
We thus have the point A' in the figure of Art. 295.
Let 6 increase from 0° to cos~^ ( " ) • Thus 1—e cos
increases algebraically from — (e — 1) to — 0,
i.e. — increases algebraically from — (e — 1) to — 0,
i.e. r decreases algebraically from = to — cc .
For these values of 6 the radius vector is therefore
negative and increases in numerical length from to oo .
THE POLAR EQUATION, FOCUS BEING POLE. 309
We thus have the portion A'P^R oo of the curve. For
this portion r is negative.
If 6 be very slightly greater than cos~^ - , then cos 6 is
slightly less than - , so that 1 -e cos is small and positive,
and therefore r is very great and is positive. Hence, as 6
increases through the angle cos~^ - , the value of r changes
from — cx) to + 00 .
As 6 increases from cos~^— to tt, 1— ecos^ increases
e
from to 1 + e and hence r decreases from oo to q .
1 +e
Now is < — - . Hence the point A, which corresponds
to 6^ = TT, is such that SA < SA'.
For values of 9 between cos~^- and tt we therefore
e
have the portion, ao RPA^ of the curve. For this portion
r is positive.
As increases from tt to 27r — cos~^ - , e cos 9 increases
e
from — e to 1, so that 1 — ecos^ decreases from 1 +e to 0,
and therefore r increases from — to oo . Corresponding
to these values of 6 we have the portion AL'R^ oo of the
curve, for which r is positive.
Finally, as increases from Stt-cos"^- to 27r, ecos^
e
increases from 1 to e, so that 1 — e cos 6 decreases algebraic-
ally from to 1 — e, i.e. - is negative and increases
r
numerically from to e— 1, and therefore r is negative and
decreases from go to . Corresponding to these values
of 6 we have the portion, oo R{A\ of the curve. For this
portion r is negative.
310 COORDINATE GEOMETRY.
r is therefore always positive for the right-hand branch
of the curve and negative for the left-hand branch.
It will be noted that the curve is described in the order
A'P^R' 00 00 RPAL'R^ oo oo R^A'.
338. In Case III. of the last article, let any straight line be
dravm through S to meet the nearer branch in ^, and the further
branch in q.
The vectorial angle of p is XS-p, and we have
I
^~l-e cos XS;p '
The vectorial angle of q is not XSq but the angle that qS produced
makes with SX, i.e. it is XSq^ir. Also for the point q the radius
vector is negative so that the relation (1) of Art. 335 gives, for the
point a,
^ l-ecos{XSq-f^7r) 1 + ecosXSq'
*-^* ^^^~l + ecosXSq'
This is the relation connecting the distance, Sq, of any point on
the further branch of the hyperbola with the angle XSq that it makes
with the initial line.
339. Equation to the directrices.
Considering the figure of Art. 295, the numerical values
of the distances SZ and SZ' are - and - + 2CZ,
e e
I.e. - and - + 2
e e e(e2-l)'
since ^^=-e-^^)' [Art. 300.]
The equations to the two directrices are therefore
r cos o = — ,
e
. rl 21 -] le^+1
and r cos c^ = — - + — — - — — = — - .
\je e(e^ — 1)J e e^—l
The same equations would be found to hold in the case
of the ellipse.
POLAR EQUATION TO A CONIC. 311
340. Equation to the asymptotes.
The perpendicular distance from S upon an asymptote
(Fig., Art. 315)
= OS sin ACK^ = as . . ^ = h.
s/a" + }p
Also the asymptote CQ makes an angle cos~^ — with the
axis. The perpendicular on it from S therefore makes an
angle ^ + cos-i - .
Hence, by Art. 88, the polar equation to the asymptote
CQ is
h = r cos ^ — 9 — cos~^ - = r sin 6 — cos~^ - .
The polar equation to the other asymptote is similarly
h — r cos 6- y-^ cos "^ - ) = — r sin {d + cos~* - j .
341. Ex. 1. In any conic, prove that
(1) the sum of the reciprocals of the segments of any focal chord
is constant, and
(2) the sum of the reciprocals of two perpendicular focal chords is
constant.
Let PSP' be any focal chord, and let the vectorial angle of P be a,
so that the vectorial angle of P' is ir + a.
(1) By equation (1)
of Art. 335, we have
^p=l ecosa,
and
A =1
SP'
- e cos (tt + a) = 1 + e (
Hence
SP^SP'~ '
so that
112
SP'^ SP'~ I'
The semi-latus-rectum is therefore the harmonic mean between
the segments of any focal chord.
.312 COORDINATE GEOMETRY.
(2) Let QSQ' be the focal chord perpendicular to PSP', so that the
vectorial angles of Q and Q' are - + a and -^ + a. We then have
7^7^ = 1 -e cos ( - + a | = l + esina,
and — - = l-ecos (-|^ + aj = l + ecos(^ + aj=l-esina.
Hence
7 7 27
PF'=SP + SP' = , +
1- e cos a 1 + e cos a 1 - e^ cos^ a '
Z Z 21
^^ ^ ^ l + esma 1-e sin a 1-e^ sm^ a
Therefore
1 1 1-e^ cos^ a 1 - e^ sin^ a _2-e^
PP''^'QQ'^ 2Z "^ 21 ~ ^2F'
and is therefore the same for all such pairs of chords.
Ex. 2. Prove that the locus of the middle jioints of focal chords of
a conic section is a conic section.
Let PSQ be any chord, the angle PSX being d, so that
I
SP
and SQ=
Let .
r and d.
1-e cos d '
I I
1-e cos [it + 6) 1 + e cos '
Let iJ be the middle point ot PQ, and let its polar coordinates be
Then .=SP-BP=SP_«^+^= «^-««
^ Ll-«cos^ 1 + e COS ^J 1-
2
ecos^
e'-^ cos 20'
i.e. r^-eVcos^^rrZe. ?'cos0.
Transforming to Cartesian coordinates this equation becomes
x^ + y^- e^x^ =lex (1) .
If the original conic be a parabola, we have e = l, and equation (1)
becomes y^ = lx, so that the locus is a parabola whose vertex is S and
latus-rectum I.
If e be not equal to unity, equation (1) may be written in the form
^ (i-')[-ir^J+'/=4-(^j
and therefore represents an ellipse or a hyperbola according as the
original conic is an ellipse or a hyperbola.
POLAR EQUATION TO THE TANGENT. 313
342. To find the 'polar equation of the tangent at any
point P of the conic section - — \~e cos B.
Let P be the point (rj, a), and let Q be another point
on the curve, whose coordinates are (r^, /?), so that we have
1 — ecosa (1),
T
and - = 1 - e cos i3 (2).
By Art. 89, the polar equation of the line PQ is
sin(;g-a) _ sin(6>-a) sin (j8 - 6)
r r^ ^1 '
By means of equations (1) and (2) this equation becomes
- sin (^ - a) = sin (^ — a) {1 — e cos /3} + sin {(i-6){\—e cos a}
= {sin(^ -a) + sin {(3-0)} -e {sin (^ -a) cos^ + sin (/3-^) cos a}
^ . B-a 20-a-B
= 2 sm ^— ^r — cos
— e{(sin^ cos a - cos ^ sin a) cos ^ + (sin /5 cos ^— cos ^ sin ^) cos a}
2 sin ^—^ — cos [B ~ J — e cos sin ifi - a),
^^_^A_ecos^ ..,(3).
I B-a //,a + y8
».e. - = sec — - — cos ' "
r A
This is the equation to the straight line joining two
points, P and Q^ on the curve whose vectorial angles, a and
^, are given.
To obtain the equation of the tangent at P we take Q
indefinitely close to P, i.e. we put /5 = a, and the equation
(3) then becomes
- =cos (^ — a) — ecos^ (4).
This is the required equation to the tangent at the
point a.
314 COOKDINATE GEOMETKY.
343. If we assume a suitable form for the equation to the
joining chord we can more easily obtain the required equation.
Let the required equation be
-=Lcos(^-7) -ecos^ (1).
[On transformation to Cartesian coordinates this equation is
easily seen to represent a straight line ; also since it contains two
arbitrary constants, L and 7, it can be made to pass through any two
points.]
If it pass through the point {r^^ a), we have
1 - e cos a =— =Xi cos (a - 7) - e cos a,
i.e. I, cos (a -7) = ! (2).
Similarly, if it pass through the point {r^, /3) on the curve, we have
Lcos(/3-7) = l (3).
Solving these, we have, [since a and ^ are not equal]
Substituting this value m (3), we obtam i = sec— ^ .
The equation (1) is then
(-^0-
-=sec— — COS ( d ^ ) -ecos^.
As in the last article, the equation to the tangent at the point a is
then
I
-=QOS{d-a)-eGosd.
r
^344. To find the polar equation of the polar of any
point (r^ , ^1) with respect to the conic section ~ — \—e cos 0.
Let the tangents at the points v^hose vectorial angles
are a and /? meet in the point (r^, ^1).
The coordinates r^ and 61 must therefore satisfy equation
(4) of Art. 342, so that
— — cos(^i — a)-ecos^i (1).
Similarly,
- = cos{6^- 13) -ecose^ (2).
• POLAR EQUATION TO THE POLAR. 315
Subtracting (2) from (1), we have
cos ($1 — a) = cos (^1 — /?),
and therefore
^1 - a = — (^1 — /?), [since a and /? are not equal],
2
Substituting this value in (1), we have
t.e
= ^1 (3).
— = COS \ — ^r- ay —e cos d-i .
n [ 2 )
%.e. cos ^-^r — = — hecos^, (4).
Also, by equation (3) of Art. 342, the equation of the
line joining the points a and fi is
- + e cos d = sec -— — cos ' "
T 2
( - + e cos ij \ cos — - — = cos (6^ — ^ J ,
(- + ecos^ j f- + ecos^ij = cos(^-^i) (5).
%,e.
This therefore is the required polar equation to the polar
of the point (r^, 6-^.
^345. To find the equation to the normal at the point
whose vectorial angle is a.
The equation to the tangent at the point a is
- = cos (0 — a) — e cos 0,
i.e., in Cartesian coordinates,
a? (cos a — e) + yBina = l (1).
Let the equation to the normal be
^cos^ + ^sin^ = - (2),
i.e. Ax + Bi/ = l (3).
316 COORDINATE GEOMETRY.
Since (1) and (3) are perpendicular, we have
A (cos a-e) + Bs>ma = (4).
Since (2) goes through the point ( ^"^ , a j we have
A cos a + ^ sin a = 1 — e cos a (5).
Solving (4) and (5), we have
1— ecosa , „ (1 — ecosa) (e — cosa)
A = , and B = /^— .
e e sm a
The equation (2) then becomes
Zesina
sm a cos + {e — cos a) sm =
r(l — e cos a) '
e sin a I
I.e. sui(0 — a)~esuid = — ^ — .-.
^ ^ l—e cos a r
346. If the axis of the conic be inclined at an angle y to the
initial line, so that the equation to the conic is
- = 1-6 cos (^-7),
r
the equation to the tangent at the point a is obtained by substituting
a - 7 and ^ - 7 for a and 6 in the equation of Art. 342.
The tangent is therefore
-=e cos {d-a)-e cos [d - 7).
The equation of the line joining the two points a and p is, by the
same article,
I B-a
- = sec^-^r- cos
r 2
The equation to the polar of the point {i\ , 6^) is, by Art. 344,
j- + gcos(^-7)l j- + e cos (^1-7)1 =cos{d-d{).
Also the equation to the normal at the point a
■ ,« V . , ^x-> elsm(a-y)
r {e sm ^ - 7 + sin a -6)} = , , \ •
*- \ 1/ \ 'i l-ecos(a-7)
347. Ex. 1. If the tangents at any two points P and Q of a
conic meet in a point T, and if the straight line PQ meet the directrix
corresponding to S in a point K, then the angle KST is a right angle.
POLAR EQUATION. EXAMPLES. 3l7
If the vectorial angles of P and Q be a and /3 respectively, the
equation to PQ is, by equation (3) of Art. 342,
I ^-
- = sec^
r 2
^cosT^-'^Vgcos^ (1).
Also the equation to the directrix is, by Art. 339,
-= -ecos^ (2).
r
If we solve the equations (1) and (2), we shall obtain the polar
coordinates of K.
But, by subtracting (2) from (1), we have
so that SK bisects the exterior angle between SP and SQ.
Also, by equation (3) of Art. 344, we have the vectorial angle of T
equal to ^— , i.e. L TSX=
2 ' 2
Hence Z KST= L KSX - z TSX='^ .
Ex. 2. S is the focus and P and Q tioo points on a conic such that
the angle PSQ is constant and equal to 25; prove that
(1) the locus of the intersection of tangents at P and Q is a conic
section ivhose focus is S,
and (2) the line PQ always touches a conic ivhose focus is S.
(1) Let the vectorial angles of P and Q be respectively 7 + 5 and
7 - S, where 7 is variable.
By equation (4) of Art. 342, the tangents at P and Q are therefore
- = cos(6*-7-5)-ecos^ (1),
and - = cos(6'-7 + 5)-ecos^ .(2).
If, between these two equations, we eliminate the variable quantity
7, we shall have the locus of the point of intersection of the two
tangents.
Subtracting (2) from (1), we have
cos {e--y-8) = cos (^ - 7 + 8).
Hence, (since 5 is not zero) we have 7=^.
318 COORDINATE GEOMETRY.
Substituting for 7 in (1), we have
-=cos5-gcos^,
r
Isead ^
i e. =1 - esec ocos^.
r
Hence the required locus is a conic whose focus is 8, whose latus
rectum is 21 sec 5, and whose eccentricity is e sec 5.
It is therefore an ellipse, parabola, or hyperbola, according as
e sec 5 is < = >1, i.e. according as cos 8> = <:e.
(2) The equation to PQ is, by equation (3) of Art. 342,
- = sec5cos(^- 7) -ecos^,
T
i,e. =cos(&-7) -ecos dcos^ (3).
T
Comparing this with equation (4) of Art. 342, we see that it always
touches a conic whose latus rectum is 21 cos 5 and whose eccentricity
is ecosS.
Also the directrix is in each case the same as that of the original
^ , ^, Z sec 5 - i cos 5 ^ ^ I
conic. For both r and are equal to - .
e sec 8 e cos 5 e
Ex. 3. A circle passes through the focus S of a conic and meets it
in four points ivhose distances from S are r^, ?2, r^, and r^. Prove that
dH^
(1) r-^r^r^r^ = —^ , lohere 21 and e are the latus rectum and
eccentricity of the conic, and d is the diameter of the circle,
. /«x 1 1 1 1 2
and (2) - + - + - + - = 7.
J-i r^ rg 7-4 I
Take the focus as pole, and the axis of the conic as initial line, so
that its equation is
- = l-ecos^ (1).
If the diameter of the circle, which passes through S, be inclined
at an angle 7 to the axis, its equation is, by Art. 172,
r=dcos{d-y) (2).
If, between (1) and (2), we eliminate 6, we shall have an equation
in r, whose roots are ?i, r^, r^, and r^.
r — l I (r — 1\^
From (1) we have cos 6= , and hence sin ^= \/ ^~\ ) ♦
and then (2) gives
r=d cos 7 cos 9 + d sin 7 sin 6,
i.e. {er^ -dco8y{r - l)}^=d^ sin2 7 [eV _ (^ _ 1^2^^
i.e. eV- 2edcos7 . r3+r2 {d^ + 2eldGosy- e2d2sin27) - 2ld^r+dH^=sO.
POLAR EQUATION. EXAMPLES. 319
Hence, by Art. 2, we have
'V2^3^4 = -^ (%
and »'2''3''4 + r3r^»i + ?4rir2 + rir2r3=-^ (4).
Dividing (4) by (3), we have
11112
-+-+- + - = 7. •
rj ra r. r^ I
EXAMPLES. XXXIX.
1. In a parabola, prove that the length of a focal chord which is
incHned at 30° to the axis is four times the length of the latus-rectum.
The tangents at two points, P and Q, of a conic meet in T, and S
is the focus ; prove that
2. if the conic be a parabola, then ST^=SP . SQ.
3. if the conic be central, then — - -^7^;^ = —^ , where b is
the semi-minor axis.
4. The vectorial angle of T is the semi-sum of the vectorial
angles of P and Q.
Hence, by reference to Art. 338, prove that, if P and Q be on
different branches of a hyperbola, then ST bisects the supplement of
the angle PSQ, and that in other cases, whatever be the conic, ST
bisects the angle PSQ.
5. A straight line drawn through the common focus *S of a
number of conies meets them in the points Pj, P^, ... ; on it is taken
a point Q such that the reciprocal of SQ is equal to the sum of the
reciprocals of /SPj, SP^,.... Prove that the locus of Q is a conic
section whose focus is 0, and shew that the reciprocal of its latus-
rectum is equal to the sum of the reciprocals of the latera recta of the
given conies.
6. Prove that perpendicular focal chords of a rectangular hyper-
bola are equal.
7. PSP' and QSQ' are two perpendicular focal chords of a conic ;
prove that ^^ + ^-^-^-^ is constant.
8. Shew that the length of any focal chord of a conic is a third
proportional to the transverse axis and the diameter parallel to the
chord.
9. If a straight line drawn through the focus S oi a. hyperbola,
parallel to an asymptote, meet the curve in P, prove that SP is one
quarter of the latus rectum.
320 COORDINATE GEOMETRY. tExS.
10 Prove that the equations - = l-ecos6 and -=-ecos0-l
r r
represent the same conic.
11. Two conies have a common focus; prove that two of their
common chords pass through the intersection of their directrices.
12. P is any point on a conic, whose focus is S, and a straight
line is drawn through 5^ at a given angle with SP to meet the tangent
at P in T ; prove that the locus of T is a conic whose focus and
directrix are the same as those of the original conic.
13. If a chord of a conic section subtend a constant angle at the
focus, prove that the locus of the point where it meets the internal
bisector of the angle 2a is the conic section
ZC0S5 ^ r. n
= l-e cos 8 COS 6.
r
14. Two conic sections have a common focus about which one of
them is turned ; prove that the common chord is always a tangent to
another conic, having the same focus, and whose eccentricity is the
ratio of the eccentricities of the given conies.
15. Two ellipses have a common focus ; two radii vectores, one to
each ellipse, are drawn from the focus at right angles to one another
and tangents are drawn at their extremities ; prove that these tangents
meet on a fixed conic, and find when it is a parabola.
16. Prove that the sum of the distances from the focus of the
points in which a conic is intersected by any circle, whose centre is at
a fixed point on the transverse axis, is constant.
17. Shew that the equation to the circle circumscribing the triangle
2a
formed by the three tangents to the parabola r = z drawn at
•^ 1 - cos d
the points whose vectorial angles are a, j3, and y, is
a 8 y . fa+B+y
r=a cosec - cosec ~ cosec ^ sm '
-)
2 2 2 V 2
and hence that it always passes through the focus.
18. If tangents be drawn to the same parabola at points whose
vectorial angles are a, /3, y, and 8, shew that the centres of the circles
circumscribing the four triangles formed by these four Hnes all lie on
the circle whose equation is
a B y 8 r^ a + ^ + y + S-
r= - a cosec ^ cosec ^- cosec ^ cosec - cos ' ^
z A d a
fg a + P+y + 8 -\
19. The circle circumscribing the triangle formed by three tangents
to a parabola is drawn; prove that the tangent to it at the focus
makes with the axis an angle equal to the sum of the angles made
with the axis by the three tangents.
XXXIX.] POLAR EQUATION. EXAMPLES. 321
20. Shew that the equation to the circle, which passes through
the focus and touches the curve - = 1 - ecos 6 at the point ^ = a, is
ril-e cos af = l cos {d -a) - el cos {9 - 2a).
21. A given circle, whose centre is on the axis of a parabola,
passes through the focus S and is cut in four points A, B, C, and D by
any conic, of given latus-rectum, having S as focus and a tangent to-
the parabola for directrix ; prove that the sum of the distances of the
points A, B, G, and D from S is constant.
22. Prove that the locus of the vertices of all parabolas that can be
drawn touching a given circle of radius a and having a fixed point on
the circumference as focus is r=2acos^-, the fixed point being the
pole and the diameter through it the initial line.
23. Two conic sections have the same focus and directrix. Shew
that any tangent from the outer curve to the inner one subtends a
constant angle at the focus.
.24. Two equal ellipses, of eccentricity e, are placed with their
axes at right angles and they have one focus S in common ; if PQ be
g
a common tangent, shew that the angle PSQ is equal to 2 sin-^ —p, •
25. Prove that the two conies — ^l-e^cos^ and - = l-e2Cos(^-a)
will touch one another, if
Zi2 (1 _ e^^) + 1^ (1 - e^^) + ^l^l^e-^e^ cos a = 0.
26. An ellipse and a hyperbola have the same focus S and
intersect in four real points, two on each branch of the hyperbola ; if
rj and r^ be the distances from S of the two points of intersection on
the nearer branch, and r^ and r^^ be those of the two points on the
further branch, and if I and V be the semi-latera-recta of the two
conies, prove that
«+r)(Ui)+(z-.)(i + i)=4.
[Make use of Art. 338.]
a
27. If the normals at three points of the parabola r=a cosec^-,
whose vectorial angles are a, /3, and 7, meet in a point whose vectorial
angle is 5, prove that 25=a + /3 + 7-7r.
L. 21
CHAPTER XV.
GENERAL EQUATION OF THE SECOND DEGREE.
TRACING OF CURVES.
348. Particular cases of Conic Sections. The
general definition of a Conic Section in Art. 196 was that
it is the locus of a point P which moves so that its distance
from a given point S is in a constant ratio to its perpen-
dicular distance FM from a given straight line ZK.
When S does not lie on the straight line ZK, we have
found that the locus is an ellipse, a parabola, or a hyperbola
according as the eccentricity e is <= or > 1.
The Circle is a sub-case of the Ellipse. For the
equation of Art. 139 is the same as the equation (6) of
Art. 247 when h^ = a^, i.e. when e = 0. In this case
The Circle is therefore a
CS=0, and SZ=:--ae = oo
e
Conic Section, whose eccentricity is zero, and whose direc-
trix is at an infinite distance.
Next, let S lie on the straight line ZK, so that S and Z
coincide.
In this case, since
SF=e.FM,
we have
. „^,, FM 1
/oF e
If e > 1, then F lies on one or
other of the two straight lines SU
and SU' inclined to KK' at an angle
t\H
K
^
^-^
IVI
^
zp^v
t
\
^
X
K'
^
\y*
^'-■©-
GENERAL EQUATION OF THE SECOND DEGREE. 323
If e = 1, then PSM is a right angle, and the locus
becomes two coincident straight lines coinciding with SX.
If e< 1, the z PSM is imaginary, and the locus consists
of two imaginary straight lines.
If, again, both KK' and S be at infinity and S be on
KK\ the lines SU and SU' of the previous figure will be
two straight lines meeting at infinity, i.e. will be two
parallel straight lines.
Finally, it may happen that the axes of an ellipse may
both be zero, so that it reduces to a point.
Under the head of a conic section we must therefore
include :
(1) An Ellipse (including a circle and a point).
(2) A Parabola.
(3) A Hyperbola.
(4) Two straight lines, real or imaginary, inter-
secting, coincident, or parallel.
349. To shew that the general equation of the second
degree
aoi? + 2hxy + hy^ + 2gx + 2/?/ + c = (1)
always represents a conic section.
Let the axes of coordinates be turned through an angle
6, so that, as in Art. 129, we substitute for x and y the
quantities x cos 6 — y sin 6 and x sin 6 + y cos 9 respec-
tively.
The equation (1) then becomes
a (x cos — y sin Oy + 2h (x cos — y sin 0) {x sin + y cos 6)
+ h(x sin + y cos Oy + 2g (x cos — y sin 6)
+ 2/{x sin 6 + y cos 6) + c = 0,
i. e. x^ (a cos^ + 2h cos sin + b sin^ 0)
+ 2xy {h (cos^ - sin^ 0)-(a — h) cos 6 sin 6]
+ 2/^ {a sin^ 6 - 2h cos OsinB + h cos^ 6) + 2x (g cos +/sin 0)
+ 2y{fcosO-gsine) + c = (2) .
21—2
324 COORDINATE GEOMETRY.
Now choose the angle 6 so that the coefficient of xy in
this equation may vanish,
i. e. so that h (cos^ 6 — sin^ 0) — (a — h) sin 6 cos ^,
i. e. 2h cos 20 = (a — b) sin 20,
2h
i. e. so that tan 20 = -, .
a
Whatever be the values of a, b, and h, there is always
a value of satisfying this equation and such that it lies
between — 45° and + 45°. The values of sin and cos are
therefore known.
On substituting their values in (2), let it become
Ax'' + JBy'' + 2Gx + 2Fy + c = (3).
First, let neither A nor B be zero.
The equation (3) may then be written in the form
Gy „/ i<^V ^' ^'
,/ Gy „/ ry g' r'
A[x^-)^B[y^-j^)=-^^^-c.
— -7 , ~ n)
The equation becomes
G^ F^
Ax^ + Bf = -j +-^-c = K (say) (4),
B
K ' K
B
i-e. Tr^^,-^ • (5).
K K
If — and -^ be both positive, the equation represents an
ellipse. (Art. 247.)
K K
If -J and — be one positive and the other negative, it
An
represents a hyperbola (Art. 295). If they be both
negative, the locus is an imaginary ellipse.
If X be zero, then (4) represents two straight lines,
which are real or imaginary according as A and B have
opposite or the same signs.
CENTRE OF A CONIC SECTION. 325
Secondly^ let either J. or ^ be zero, and let it be ^.
Then (3) can be written in the form
G F^
-(-3
F\2
^ "^ 2G 2BG
y
0.
Transform the origin to the point whose coordinates
are
__c_ J^ _F
~2G'^2M' ~B
This equation then becomes
By^+2Gx^0,
2G
^.e. r^--^^'
which represents a parabola, (Art. 197.)
If, in addition to A being zero, we also have G zero, the
equation (3) becomes
By^ + 2Fy + c = 0,
F F^ G
'^^B^^\/w-r
and this represents two parallel straight lines, real or
imaginary.
Thus in every case the general equation represents one
of the conic sections enumerated in Art. 348.
350. Centre of a Conic Section. Def. The
centre of a conic section is a point such that all chords of
the conic which pass through it are bisected there.
When the equation to the conic is in the form
ax^ + 2hxy + by'^ + g-0 (1),
the origin is the centre.
For let (x, y) be any point on (1), so that we have
aa;'2 + 2Aa;y + 62/'' + c = (2).
This equation may be written in the form
a (- ^f + 2A (- a;') (- y') + 6 (- yj + c = 0,
and hence shews that the point (— a:', — y'^ also lies on (1).
326 COORDINATE GEOMETRY.
But the points {x\ y) and {—x\ —y) lie on the same
straight line through the origin, and are at equal distances
from the origin.
The chord of the conic which passes through the origin
and any point {x, y) of the curve is therefore bisected at
the origin.
The origin is therefore the centre.
351. When the equation to the conic is given in the
form
aa? + 27t£C2/ + hy"^ + 2gx + 2fy + c = (1),
the origin is the centre only when both f and g are zero.
For, if the origin be the centre, then corresponding
to each point {x, y) on (1), there must be also a point
(—a;', —y') lying on the curve.
Hence we must have
ax^ + 'i.hxy + hy'^ + 2gx +'ify'-\-c = (2),
and ax'^ + Ihx'y' + hy'^^ - 2gx' -2fy' + c=^0 (3).
Subtracting (3) from (2), we have
gx +fy' - 0.
This relation is to be true for all the points {og\ y')
which lie on the curve (1). But this can only be the case
when g — and f= 0.
352. To obtain the coordinates of the centre of the
conic given hy the general equation, and to obtain the
equation to the curve referred to axes through the centre
parallel to the original axes.
Transform the origin to the point (S, y)j so that for x
and y we have to substitute x + x and y + y. The equation
then becomes
a{x + xf + 2h{x + x){y + y) + b (y + yY + 2g(x + x)
i. e. ax^ + 2hxy + by^ + 2x (ax + hy + g) + 2y (Jix + by +f)
+ ax^'}-2hxy + by'^ + 2gx + 2fy + c=^0 (2).
EQUATION REFERKED TO THE CENTRE. 327
If the point (x, y) be the centre of the conic section, the
coefficients of x and y in the equation (2) must vanish, so
that we have
ax + hy + g = (3),
and hx + hy +f = (4).
Solving (3) and (4), we have, in general,
With these values the constant term in (2)
= a^ + 2hxy + hy^ + 2gx + Ify + c
= X {ax + hy + g) + y (hx + hy +/) + gx+/y + c
= gx + fy + c (6),
by equations (3) and (4),
abc + 2 fgh — af^ — bi/^ — ch^ , ,. _,
= ^^ — — , by equations (5),
A
~ab^^^'
where A is the discriminant of the given general equation
(Art. 118).
The equation (2) can therefore be written in the form
aaf + 2hxy + by^ + -7 — ^ = 0.
This is the required equation referred to the new axes
through the centre.
Ex. Find the centre of the conic section
2x^ - 5xy - Sy^ ~x-iy + Q = 0,
and its equation when transformed to the centre.
The centre is given by the equations 2x~^y-^ = 0, and
-^x-By -2 = 0, sothatS=-y, and ?/= -|.
The equation referred to the centre is then
2x^-5xy-Sy^ + c' = 0,
where c'= -|. S-2.?7 + 6 = ^+f + 6 = 7. (Art. 352.)
The required equation is thus
2x^-5xy-3y^ + 7 = 0.
328 COORDINATE GEOMETRY.
353. Sometimes the equations (3) and (4) of the last
article do not give suitable values for x and y.
For, if ah — 1^ be zero, the values of x and y in (5) are
both infinite. When ah — h^ is zero, the conic section is a
parabola.
The centre of a parabola is therefore at infinity.
Again, if - = - = ^ , the result (5) of the last article is
of the form ~ and the equations (3) and (4) reduce to the
same equation, viz.,
ax + hy + g = 0.
We then have only one equation to determine the
centre, and there is therefore an infinite number of centres
all lying on the straight line
ax -{-hy -{-g — O.
In this case the conic section consists of a pair of
parallel straight lines, both parallel to the line of centres.
354. The student who is acquainted with the Dif-
ferential Calculus will observe, from equations (3) and (4)
of Art. 352, that the coordinates of the centre satisfy the
equations that are obtained by difierentiating, with regard
to x and y, the original equation of the conic section.
It will also be observed that the coefficients of x, y, and
unity in the equations (3), (4), and (6) of Art. 352 are the
quantities (in the order in which they occur) which make
up the determinant of Art. 118.
This determinant being easy to write down, the student
may thence recollect the equations for the centre and the
value of c.
The reason why this relation holds will appear from the
next article.
355. Ex. Find the condition that the general equation of the
second degree may represent two straight lines.
The centre {x, y) of the conic is given by
ax-\-hy-\-g = (1),
and hx + by+f=0 (2).
EQUATION TO THE ASYMPTOTES. 329
Also, if it be transformed to the centre as origin, the equation
becomes
ax^+2hx7j + hif + c' = (3),
where c' = gx+fy + c.
Now the equation (3) represents two straight lines if c' be zero,
i.e. if gx+fy + c—Q (4).
The equation therefore represents two straight lines if the relations
(1), (2), and (4) be simultaneously true.
Eliminating the quantities ^ and y from these equations, we have,
by Art. 12,
a, h, g
h, b, f =0.
This is the condition found in Art. 118.
356. To find the equation to the asymptotes of the conic
section given hy the general equation of the second degree.
Let the equation be
a'3(? + 2hxy + hy'^ + 2gx + ^fy + c-O... (1).
Since the equation to the asymptotes has been shewn to
differ from the equation to the curve only in its constant
term, the required equation must be
ax^ + '2hxy + hy^ + 2gx + 2fy + c + A, = (2).
Also (2) is to be a pair of straight lines.
Hence
ah (c + X) + ^fgh - af - 6/ - (c + X) K- = 0. (Art. 116.)
ahG + 2fgh-af^-hg^-c¥ A
ineretore A = — ■ ; ,„ — = -^ — j-^.
ao — a^ ab — A"
The required equation to the asymptotes is therefore
ax^ + 2hxy + hy^ + 2gx + 2/y + c j — 7^= . . .(2).
Cor. Since the equation to the hyperbola, which is
conjugate to a given hyperbola, differs as much from the
equation to the common asymptotes as the original equation
does, it follows that the equation to the hyperbola, which is
conjugate to the hyperbola (1), is
ax^ + 2hxy + hy"^ + 2gx + 2fy + c— 2 - — y^ = 0.
330 COORDINATE GEOMETRY.
357. To determine hy an examination of the general
equation what hind of conic section it represents.
[On applying the method of Art. 313 to the ellipse and
parabola, it would be found that the asymptotes of the
ellipse are imaginary, and that a parabola only has one
asymptote, which is at an infinite distance and perpen-
dicular to its axis.]
The straight lines as(? + Ihxy + hy'^ = (1)
are parallel to the lines (2) of the last article, and hence
represent straight lines parallel to the asymptotes.
Now the equation (1) represents real, coincident, or
imaginary straight lines according as J^ is >= or <a6,
i.e. the asymptotes are real, coincident, or imaginary,
according as A^ > = or < ab^ i. e. the conic section is a hyper-
bola, parabola, or ellipse, according as 1^ > = or < ah.
Again, the lines (1) are at right angles, i.e. the curve is
a rectangular hyperbola, if a ^-h — ^.
Also, by Art. 143, the general equation represents a
circle if a = h, and h = 0.
Finally, by Art. 116, the equation represents a pair of
straight lines if A = ; also these straight lines are parallel
if the terms of the second degree form a perfect square, i.e.
if h^ = ah.
358. The results for the general equation
ao(? + 2hxy + hy^ + 2gx + 2fy + c =
are collected in the following table, the axes of coordinates
being rectangular.
Curve.
Ellipse.
Parabola.
Hyperbola.
Circle.
Rectangular hyperbola.
Two straight lines, real or
imaginary.
Two parallel straight lines.
Condition.
h^ < ah.
A^ = ah.
W- > ah.
a = h, and A = 0.
a + h = 0.
A-O,
i.e.
ahc+2fgh-af^-hg''-ch''^0.
A = 0, and h^ ~ ah.
EXAMPLES. 331
If the axes of coordinates be oblique, the lines (1) of Art. 356 are
at right angles if a + 6- 2/icos a;=0 (Art. 93); so that the conic
section is a rectangular hyperbola ii a + b-2h cos w = 0.
Also, by Art. 175, the conic section is a circle if & = a and
h= a cos la.
The conditions for the other cases in the previous article are the
same for both oblique and rectangular axes.
EXAMPLES. XL.
What conies do the following equations represent? "When
possible, find their centres, and also their equations referred to the
centre.
1. 12x^--2Sxy + 10y^-25x + 2Qy = U.
2. lSx^-18xy + S7y^ + 2x + Uy -2 = 0.
3. y^-2^3xy + 3x^ + 6x-4:y + 5 = 0.
4. 2x^-72xy + 2Sy^-4:X-2Sy-^8 = 0.
5. 6x^-.5xy-Qy'^ + 14:X + 5y + 4:=0.
6. Sx^-8xy-Sy^ + 10x-lSy + 8 = 0.
Find the asymptotes of the following hyperbolas and also the
equations to their conjugate hyperbolas.
7. 8x^-\-10xy-Sy^-2x + 4:y = 2. 8. y^-xy -2x'^-5y + x-6 = 0.
9. 55x^-120xy + 20y^ + G4:X-48y = Q.
10. 19a;2 + 24:xy + %/- 22a; - 6?/ = 0.
11. If (S, y) be the centre of the conic section
f{x, y) = ax^ + 2hxy + hij'^ + 2gx 4- 2fy + c = 0,
prove that the equation to the asymptotes is/ (a;, y)=f[x, y).
If t be a variable quantity, find the locus of the point {x, y) when
12. a; = a f i + - j and ?/=« ( *--)
13. x = at + U' and y = bt + at^.
14. x = l + t + t'^ and y = l-t + t\
If ^ be a variable angle, find the locus of the point {x, y) when
15. a; = a tan (^ + a) and y = btsin{d + ^).
16. a; = acos(^ + a) and ?/ = &cos (^ + i8).
What are represented by the equations
17. {x-y)^ + (x-a)^=0. 18. xy + a^=a{x + 2j).
332
COOKDINATE GEOMETRY.
[Exs. XL.]
19. x^-7f = {y-a){x'^- 2/2).
20. x^ + y^-xy{x + y) + a^{y-x) = Q.
22. x^ + y'^ + {x + y) {xy-ax-ay) = 0.
24. {r cos 6-a){r-a cos 6) = 0.
26. r+-=3cos^ + sin0.
r
28. r (4 - 3 sin'-^ d) = 8a cos d.
21. (a;2-a2)2_ 2/4=0.
23. a;2 + a;r/ + 2/2=0.
25. rsin2^ = 2acos^.
27. -=l + cos^ + ^3sin^.
359. To trace the parabola given by the general equa-
tion of the second degree
a^ + 2hxy + b'lf' + 2^cc + yy + c = (1),
and to find its latus rectum.
First Method. Since the curve is a parabola we
have A^ = ab, so that the terms of the second degree form
a perfect square.
Put then a — c^ and 6 = yS^, so that h = a(B, and the
equation (1) becomes
{ax + Pyf + 2gx-v 2fy + c = (2).
Let the direction of the axes be changed so that the
straight line ax + Py = 0, i.e. y = — -^x, may be the new-
axis of -Z".
We have therefore to turn the axes through an angle
/8
such that tan ^ = — — , and therefore
sin = —
Va2 + ^2
and cos 6 =
sJa^ + P"'
TRACING OF PARABOLAS. 333
For X we have to substitute
JT cos — Fsin 0, i. e. ,
Ja' + ft'
and for y the quantity
JTsin^ + Tcos^, i.e. -"-- ^+1^ ^ (Art. 129.)
For ax + Py we therefore substitute Y \J{a? + y8^).
The equation (2) then becomes
T' (a? + yS') + ^jj^[9 (P^ + « Y) +/(lir - aX)] + = 0,
^.e.
<^-^)^^'(£^f^-^j (^)'
where /i = - ~^ — Bl (4)
and ^2^l=J^xff=K^- ^
I.e.
H=
{a'+0 " " a^ + yS^'
The equation (3) represents a parabola whose latus
rectum is 2 , , whose axis is parallel to the new axis
(a' + JS'f
of X, and whose vertex referred to the new axes is the
point (R, K).
360. Equation of the axis, and coordinates of the
vertex, referred to the original axes.
Since the axis of the curve is parallel to the new axis of
X, it makes an angle 9 with the old axis of x, and hence
the perpendicular on it from the origin makes an angle
90° + ^.
Also the length of this perpendicular is K.
384 COORDINATE GEOMETRY.
The equation to the axis of the parabola is therefore
X cos (90° + 6)+y sin (90° + 6) = K,
i.e. —X sin + y cos = IC,
i.e. ax^Py = Kj^TW=^-"^^ (6).
Again, the vertex is the point in which the axis (6)
meets the curve (2).
We have therefore to solve (6) and (2), i.e. (6) and
^ilT^+^^^+^-^2/ + '^ = o (7).
The solution of (6) and (7) therefore gives the required
coordinates of the vertex.
361. It was proved in Art. 224 that if PK be a
diameter of the parabola and Q V the ordinate to it drawn
through any point Q of the curve, so that ^ F is parallel to
the tangent at P, and if 6 be the angle between the diameter
P V and the tangent at P, then
^F2 = 4acosec2^.PF (1).
If QL be perpendicular to P F and QL' be perpendicular
to the tangent at P, we have
QL^QVBm.e, and ^i:' = PFsin6i,
so that (1) is QL^ - ia cosec . QL'.
Hence the square of the perpendicular distance of any
point Q on the parabola from any diameter varies as the
perpendicular distance of Q from the tangent at the end of
the diameter.
Hence, if Ax + By + C — be the equation of any
diameter and A'x + B'y + C — be the equation of the
tangent at its end, the equation to the parabola is
{Ax + By-hCf = X{A'x-\-B'y^C') (2),
where X is some constant.
Conversely, if the equation to a parabola can be reduced
to the form (2), then
Ax-\-By+G^O (3)
TRACING OF PARABOLAS. 335
is a diameter of the parabola and the axis of the parabola is
parallel to (3).
We shall apply this property in the following article.
362. To trace the parabola given hy the general equa-
tion of the second degree
ax^+ ^hxy + hy"^ + 2gx + %fy + g=^ (1).
Second Method. Since the curve is a parabola, the
terms of the second degree must form a perfect square
and A^ = ah.
Put then a — a^ and h ~ /3^, so that h = ayS, and the
equation (1) becomes
{ax + ^yf = -{2gx + 2/y + c) (2).
As in the last article the straight line ax + jSy = is a
diameter, and the axis of the parabola is therefore parallel
to it, and so its equation is of the form
ax + /3y + \ = (3).
The equation (2) may therefore be written
{ax + l3y + Xy = -{2gx+ 2fy + c) + X^+2\ (ax + fy)
= 2x{Xa-g) + 2y((3\-f) + X'-G (4).
Choose A, so that the straight lines
ax + /3y + \ = (5)
and 2x{Xa-g) + 2y(j3X-f)+X^-G = (6)
are at right angles, i.e. so that
a{Xa-g) + pH3X-/) = 0,
i.e. so that X= „ „ „ (7).
a? + ^^ ^ '
The lines (5) and (6) are now, by the last article, a
diameter and a tangent at its extremity ; also, since they
are at right angles, they must be the axis and the tangent
at the vertex.
336 COORDINATE GEOMETRY.
The equation (4) may now, by (7), be written
z e.
where PiV is the perpendicular from any point F of the
curve on the axis, and A is the vertex.
Hence the axis and tangent at the vertex are the lines
(5) and (6), where X. has the value (7), and the latus rectum
= 2
f •
(a? + 13^)
363. Ex. Trace the parabola
9a;2 - 'iixy + 16r/2 - 18a; - 101?/ + 19 = 0.
The equation is
(3a;-42/)2-18a;-101^ + 19 = (1).
First Method. Take ^x-^y = as the new axis of x, i.e. turn
the axes through an angle ^, where tan^ = |, and therefore sin^ = |
and cos^=|.
For X we therefore substitute Xcos^-Fsin^, i.e. — = ; for
5
3X + 4Y
y we put Xsin^+Fcos^, i.e. = — , and hence for ^x-Ay the
quantity -57.
The equation (1) therefore becomes
2572_i[72Z-54Y]-i[303Z+404r| + 19 = 0,
i.e. 25r2-75Z-70r+19=0 (2).
This is the equation to the curve referred to the axes OX and OY.
But (2) can be written in the form
147
T
r2_i_=3Z-^|,
i.e. (7-^)2=3Z-i| + f|=3(Z+f).
TRACING OF PARABOLAS.
337
Take a point A whose coordinates referred to OX and OY are -|
and I, and draw AL and AM parallel to OX and OY respectively.
lA
Referred to AL and AM the equation to the parabola is Y^=SX.
It is therefore a parabola, whose vertex is A, whose latua rectum is 3,
and whose axis is AL.
Second Metliod. The equation (1) can be written
(3a;-4?/ + X)2=(6\ + 18)a; + ?/{101-8X) + X2-19 (3).
Choose X so that the straight lines
Sx-4y + \=0
and (6X + 18)a; + y (101- 8X) + X2- 19 =
may be at right angles.
Hence X is given by
3 (6X + 18) - 4 (101 - 8X) = (Art. 69),
and therefore X = 7.
The equation (3) then becomes
(8a; -4:y + If = 15 {4.x + 3i/ + 2),
(^^r--^^^ <^)-
Let AL be the straight line
3a;-4^ + 7 = (5),
and blithe straight line 4a; + 3z/ + 2 = (6).
These are at right angles.
If P be any point on the parabola and FN be perpendicular to
AL, the equation (4) gives PN^=S . AN.
Hence, as in the first method, we have the parabola.
The vertex is found by solving (5) and (6) and is therefore the
point (-If, ft).
L. 22
338 COORDINATE GEOMETRY.
In drawing curves it is often advisable, as a verification, to find
whether they cut the original axes of coordinates.
Thus the points in which the given parabola cuts the axis of x
are found by putting 2/ = in the original equation. The resulting
equation is Occ^- 18a; + 19 = 0, which has imaginary roots.
The parabola does not therefore meet Ox.
Similarly it meets Oy in points given by 16?/2- 101i/ + 19 = 0, the
roots of which are nearly %\ and y\.
The values of OQ and OQ' should therefore be nearly ^^ and 6^.
364. To find the direction and magnitude of the axes
of the central conic section
aQ(? + 2Axy + hy"^ =1 (1).
First Method. We know that, when the equation to
a central conic section has no term containing xy and the
axes are rectangular, the axes of coordinates are the axes of
the curve.
Now in Art. 349 we shewed that, to get rid of the term
involving xy^ we must turn the axes through an angle 6
given by
tan2^--^ (2).
a — ^ '
The axes of the curve are therefore inclined to the axes
of coordinates at an angle 6 given by (2).
Now (2) can be written
2 tan ^ _ 2A _ 1
l-tan^^"^^6-X^'^^^'
.-. tan2^ + 2A.tan(9-l =0 (3).
This, being a quadratic equation, gives two values for ^,
which differ by a right angle, since the product of the two
values of tan ^ is — 1. Let these values be 6^ and 6^, which
are therefore the inclinations of the required axes of the
curve to the axis of x.
Again, in polar coordinates, equation (1) may be written
r"" (a cos2 e + 2h cos (9 sin ^ + 6 sin' $)=! = cos^ 6 + sin' (9,
i.e.
cos2(9 4-sin2^ 1+tan'^
^'' = _ — — —
a cos^ + 2h cos sin $ + b sin^ a + 2h tan 6 + b tan^ 6
(4)..
AXES OF A CENTRAL CONIC SECTION. 839
If in (4) we substitute either value of tan 6 derived
from (3) we obtain the length of the corresponding
semi-axis.
The directions and magnitudes of the axes are therefore
both found.
Second Method. The directions of the axes of the
conic are, as in the first method, given by
tan 2^=.-^.
a —
When referred to the axes of the conic section as the
axes of coordinates, let the equation become
- + ■^-1 rm
Since the equation (1) has become equation (5) by a
change of axes without a change of origin, we have, by
Art. 135,
1 1
-2 + ^2 = ^ + ^ (6),
and -l^^ah-h' (7).
arp-'
These two equations easily determine the semi-axes a
and /5. [For if from the square of (6) we subtract four
times equation (7) we have (-^ — 02) , and hence —^— -^^'j
1 In
hence by (6) we get —^ and — .
The difficulty of this method lies in the fact that we
cannot always easily determine to which direction for an
axis the value a belongs and to which the value yS.
If the original axes be inchned at an angle w, the equa-
tions (6) and (7) are, by Art. 137,
1 1 a + h — 2h cos w
, \ ah — //
and ~¥7yi= - 2 '
22—2
340 COORDINATE GEOMETRY.
Cor. 1. The reciprocals of the squares of the semi-
axes are, by (3) and (4), the roots of the equation
2^ - {ci + h) Z + ah -h^ = 0.
Cor. 2. From equation (4) we have
Area of an ellipse = 7ra/5 =
\lah — h^
365. Ex. 1. Trace the curve
14a;2 - 4a;2/ + 112/2- 44a; -58i/ + 71 = (1).
Since ( - 2)^ - 14 . 11 is negative, the curve is an ellipse. [Art. 358.]
By Art. 352 the centre (S, y) of the curve is given by the equations
145-2^-22=0, and -25 + 11^-29 = 0.
Hence ic= 2, and ^=3.
The equation referred to parallel axes through the centre is
therefore l^x^ - ^xy + lly'^ + c' = 0,
where c' = - 225 - 29^ + 71 = - 60,
so that the equation is
Ux^-4:xy + lly'^ = m (2).
The directions of the axes are given by
,. „„ 2h -4
*"^2^=^36=i43ri=--
2tan^ .
so that
l-tan2^ ^'
and hence 2 tan^ ^ - 3 tan - 2 = 0.
Therefore tan ^i=2, and tan 6^= -\.
Referred to polar coordinates the equation (2) is
r2 (14 cos2 ^ - 4 cos ^ sin ^ + 11 sin^ d) = 60 (cos^ 9 + sin^ 0),
l + tan2^
r''^ = 60
14-4tan6' + lltan2 6'*
When tan ^1 = 2, »'i' = 60x ^^^^=6.
When tan 6^=-^ r^^=GO x j^j:j:|^_ =4.
TRACING A CENTRAL CONIC SECTION.
341
The lengths of the semi-axes are therefore fJ6 and 2.
Hence, to draw the curve,
take the point C, whose coordi-
nates are (2, 3).
Through it draw A'CA in-
clined at an angle tan~i 2 to the
axis of X and mark off
A'C = GA=>J6.
Draw BCB' at right angles
to AC A' and take B'G=CB=:2.
The required ellipse has A A'
and BB' as its axes.
It would be found, as a veri-
fication, that the curve does not meet the original axis of x, and
that it meets the axis of y at distances from the origin equal to
about 2 and 3 J respectively.
Ex. 2. Trace the curve
x'^~Bx2j + if + 10x-10y + 21 =
1 . 1 is positive, the curve is a hyperbola.
.(1).
Since
(^7
[Art. 358.]
The centre {x, y) is given by
- 3_ ^ ^
and —x + y -5 = 0,
so that ^=-2, and ^ = 2.
The equation to the curve, referred to parallel axes through the
centre, is then
a;2 - 3a;i/ + 2/2 + 5 ( - 2) - 5 X 2 + 21 = 0,
i.e. x^-3xy + y^=-l (2).
The direction of the axes is given by
2h -3
tan 20 =
:0O,
a-b 1-1
so that 20=90° or 270°,
and hence 0^= 45° and 02=135°.
The equation (2) in polar coordinates is
r2 (cos2 0-3 cos sin + sin^ 0) = - (sin^ + cos^ 0),
l + tan2
I.e.
l-3tan0 + tan2
342
COORDINATE GEOMETRY.
When ^1 = 45°, r^^^ - ^-^— =2, so that i\ = sl'^
-2
When 6'2=135°, r^- -
2
, so that 7
■^v
-2
5
2--" > '2- 1 + 3 + 1" 5
To construct the curve take the point G whose coordinates are - 2
and 2. Through G draw a straight line AG A' inclined at 45° to the
axis of X and mark off A'C=GA = J2.
Also through A draw a straight line KAK' perpendicular to GA
and take AK=^K'A = ^^. By Art. 315, GK and GK' are then the
asymptotes.
The curve is therefore a hyperbola whose centre is G, whose
transverse axis is ^'^, and whose asymptotes are GK and GK'.
On putting a;=0 it will he found that the curve meets the axis of
y where y — Z or 7, and, on putting ?/ = 0, that it meets the axis of x
where x= - 3 or - 7.
Hence
0*3=3, 0(9' = 7, Oi? = 3, and 0R' = 1.
366. Tojind the eccentricity of the central conic section
ax^-\- 2hxy ■\-hy'^= 1 (1).
First, let 7i^ — ah be negative, so that the curve is
ECCENTEICITY OF A CONIC SECTION. 343
an ellipse, and let the equation to the ellipse, referred to
its axes, be
By the theory of Invariants (Art. 135) we have
^^+|-^ = ^ + ^ (2),
and a^^"''^"'^'' (^)'
Also, if e be the eccentricity, we have, if a be > ^,
^ = 2—-
a.
<2'
" 2-6" a' + ZS^
But, from (2) and (3), we have
Hence
ab — h^
e" _ s/(a-bf + 4:h^
2-e^ ' a + b ^^^'
This equation at once gives e^.
Secondly^ let h- — ab be positive, so that the curve is
a hyperbola, and let the equation referred to its principal
axes be
^ _ ^^ = 1
so that in this case
-2- o2 = ^* + ^'a^^-^2^2 = ^^-'^^' = -(^'-^^)•
Hence a^ — B^ = — T^ r s-nd a^B^ = -r^ j ,
' /r — ab hr — ab
so that a^ + /3^ = + x/(a^ - ^J + 4a^/3^ ^ + h?-]l ^^'^ '
344 COORDINATE GEOMETRY.
In this case, if e be the eccentricity, we have
'"^- j:^^~a''^'~~ a + b ^^'
This equation gives e^.
In each case we see that e is a root of the equation
«2 N2 ^a-bY + 4:h^
\2-e')
(a + hf '
i.e. of the equation
e^ {ab - h') + {{a-bY + ih'} (e^ - 1 ) - 0.
367. To obtain the foci of the central conic
aoi? + Ihxy + by^ — 1.
Let the direction of the axes of the conic be obtained as
in Art. 364, and let 6-^ be the inclination of the major axis
in the case of the ellipse, and the transverse axis in the case
of the hyperbola, to the axis of x.
Let r/ be the square of the radius corresponding to ^i,
and let r^ be the square of the radius corresponding to the
perpendicular direction. [In the case of the hyperbola r^
will be a negative quantity.]
The distance of the focus from the centre is ^jr^ — r^
(Arts. 247 and 295). One focus will therefore be the point
(vri^ — T^ cos ^1 , Jr^ — r^ sin ^J,
and the other will be
(- sjr^ — ri cos 6^ , - Jr^ - r/ sin 6^).
"Ex.. Find the foci of the ellipse traced in Art. 365.
2 1
Here tan 6-^ = 2, so that sin 6^=-j^ and cos ^j = -^ .
Also ri2 = 6, and 7-22=4, so that sjrf^=^2.
The coordinates of the foci referred to axes through C are therefore
FOCI OF A CONIC.
345
Their coordinates referred to the original axes OX and OY are
2±
V2
x/5'
2V2\
368. The method of obtaining the coordinates of the
focus of a parabola given by the general equation may be
exemplified by taking the example of Art. 363.
Here it was shewn that the latus rectum is equal to 3,
so that, if aS' be the focus, ^aS' is J.
It was also shewn that the coordinates of A referred to
OX and Y are — ^ and ^.
The coordinates of S referred to the same axes are
+
f and -I,
^.e.
2V and -I
Its coordinates referred to the original axes are therefore
_7_
20
W COS
i.e.
7
2 J)
Z- sin 6 and -J-pr sin + ~ cos
+
i-i and
2 5
^* a
7 3
20*5
nd 1-^3
7_
5
I. 1
5*5'
^,
100'
In Art. 393 equations will be found to give the foci of
any conic section directly, so that the conic need not first
be traced.
369. Ex. 1. Trace the curve
The equation may be written
.(l).
(^^^T-K'^^T- <^'-
Now the straight lines 3a;- 2?/ + 4 = and 2x + Sy-5 = are at
right angles. Let them be C3I and
GN, intersecting in C which is the
point (-t\, H)-
If P be any point on the curve
and PM and PN the perpendiculars
upon these lines, the lengths of P3I
and PN are
3x-2y + 4: ^ 2x + 3y-5
-^w ^^^ --^w- ■
Hence equation (2) states that
3PM2 + 2P^■2=:3,
PM^
I.e.
PN^
346 COORDINATE GEOMETRY.
The locus of P is therefore an ellipse whose semi-axes measured
along CM and CN are a/I and 1 respectively.
Ex. 2. What is represented by the equation
{x'^-a^f + iy^-a^)^=a*?
The equation may be written in the form
a;4 + t/-i - 2a2 (a;2 + ?/2) + a* = 0,
i.e. {x^ + y^f-2a^{x^ + y^)+a*=2xY,
i.e. {x^ + y^-a^)^-{^2xyf=0,
i.e. {x^ + y/^^y +y^- «^) (^^ - fj^^y +y^- ^^) = O-
The locus therefore consists of the two ellipses
x^ + /ij2xy+y^-a^=0, and x^-fj2xy + y'^-a^=0.
These ellipses are equal and their semi-axes would be found to be
asj2 + ^2 and a ^2 -^2.
The major axis of the first is inclined at an angle of 135° to the
axis of X, and that of the second at an angle of 45°.
EXAMPLES. XLI.
Trace the parabolas
1. {x-4:ij)^=51y. 2. {x-y)^=x + y + l.
3. {5x-12yf=2ax + 2day + a\
4. {4a; + 3?/ + 15)2= 5 (3a; -4?/).
5. 16a;2 + 24a;2/ + 92/2-5a;-10?/ + l = O.
6. 9x^+24^y + 16y^-4:y-x + 7 = 0.
7. 144a;2 - 120xy + 25y^ + 619a; - 272^ + 663 = 0, and find its focus.
8. 16a;2-24a;2/ + 9?/2 + 32a; + 86?/-39 = 0.
9. 4a;2-4a;2/ + 2/2-12a; + 62/ + 9 = 0.
Find the position and magnitude of the axes of the conies
10. 12a;2-12a;i/ + 7?/2=48. 11. 3a;2+2a;?/ + 32/2=8.
12. x'^-xy~6y^=Q.
Trace the following central conies.
13. a;2-2a;2/cos2a + ^2— 2a2. 14. a;2-2a;2/cosec2a-f2/^ = a2.
15. xy = a{x + y). 16. xy-y^ = a^.
17. 2/^-2a;?/ + 2a;2 + 2a;-22/ = 0. 18. x'^ + xy + y^ + x+y = l.
LEXS. XLI.] TRACING OF CONIC SECTIONS. EXAMPLES. 347
19. 2x^ + Zxii-2y'^-lx + y-2 = Q.
20. 40a;2 + 36a;?/ + 25?/2- 196a; -122?/ + 205 = 0.
21. x^-%xy + y^ + 10x-l(iy + 21=^0.
22. x'^-xy + 2^2 _ 2ax - Qay + 7a^ = 0.
23. 10a;2 - ASxy - lOif + dSx + Uy - 5^ = 0.
24. 4:x^ + 27xy + S5y^~14M-Bly-&:=0.
25. {3x-4y + a){4:X + Sy + a) = a\
26. 3 (2a; -3^ + 4)2 + 2 (3a; + 2?/ -5)2 = 78.
27. 2 (3a; -4^/ + 5)2 -3 (4a; + 3?/ -10)2 = 150.
Find the products of the semi-axes of the conies
28. 2/2-4a;i/ + 5a;2 = 2. 29. 4(3a; + 42/- 7)2 + 3 (4a;-3t/ + 9)2=3.
30. lla;2 + 16a;?/ - ?/2- 70a; -40y + 82 = 0.
Find the foci and the eccentricity of the conies
31. a;2-3a;i/ + 4aa; = 2a2. 32. 4a;?/ - 3a;2 - 2a?/ = 0.
33. 5a;2 + 6a;?/ + 5?/ + 12a; + 4?/ + 6 = 0.
34. a;2 + 4a;?/ + ?/2-2a; + 2/y-6 = 0.
35. Shew that the latus rectum of the parabola
(a2 + 62) (^^2 + ^2) ^ (5^ + ay _ ahf
is 2ab-hs/a^~+^.
36. Prove that the lengths of the semi-axes of the conic
aa;2 + 2hxy + ay^ = d
are \/ — — r and
respectively, and that their equation is a;2_y2_o_
37. Prove that the squares of the semi-axes of the conic
ax2 + 2/?a;?/ + &?/2 + 2(;a; + 2/?/ + c =
are 2A-^ { {ab -h^)(a + h^ J {a - hf + 4/i2)} ,
where A is the discriminant.
38. If X be a variable parameter, prove that the locus of the
vertices of the hyperbolas given by the equation x^-y'^ + \xy = a^ is
the curve {x^+y^)^=a^{x^-y^).
39. If the point {at^^, 2at-^) on the parabola ?/2=4aa; be called the
point t^, prove that the axis of the second parabola through the four
points ^1, t^, ^3, and t^ makes with the axis of the first an angle
oot-(*l±^±i±^A.
Prove also that if two parabolas meet in four points the distances
of the centroid of the four points from the axes are proportional to the
latera recta.
348 COORDINATE GEOMETRY. [ExS. XLI.]
40. If the product of the axes of the conic x^ + 2xy + ny'^=8 be
unity, shew that the axes of coordinates are inclined at an angle
sin-i 1.
41. Sketch the curve Qx^-7xy-5y^~4:X + lly = 2, the axes being
inclined at an angle of 30°.
42. Prove that the eccentricity of the conic given by the general
equation satisfies the relation
e^ , (a + & - 2/1 cos w)2
h 4 = - —
1 - e^ " (a6 - h^) sin^ w '
where w is the angle between the axes.
43. The axes being changed in any way, without any change of
origin, prove that in the general equation of the second degree the
P + 9^ - 2fg cos 0} ap+bg^-2fgh , A
quantities c, '' •^ . ^^ , -^— — — — =^^ , and -.-^- are
sm^w sm^w sin^w
invariants, in addition to the quantities in Art. 137.
[On making the most general substitutions of Art. 132 it is clear
that c is unaltered; proceed as in Art, 137, but introduce the condition
that the resulting expressions are equal to the product of two linear
quantities (Art. 116); the results will then follow.]
CHAPTER XVL
THE GENERAL CONIC.
370. In the present chapter we shall consider proper-
ties of conic sections which are given by the general equation
of the second degree, viz.
aoi? + '2Jnxy + hy^ + Igx + ^fy + c = (1).
For brevity, the left-hand side of this equation is often
called ^ (tc, 3/), so that the general equation to a conic is
^ (a;, y) = 0.
Similarly, <^(aj', y) denotes the value of the left-hand
side of (1) when x and y are substituted for x and y.
The equation (1) is often also written in the form aS'^O.
371. On dividing by c, the equation (1) contains five
independent constants - , - , - , - , and - .
c G G G c
To determine these five constants, we shall therefore
require five conditions. Conversely, if five independent
conditions be given, the constants can be determined.
Only one conic, or, at any rate, only a finite number of
conies, can be drawn to satisfy five independent conditions.
372. To find the equation to the tangent at any foint
{x\ y') of the conic section
cf> {x, y) = ax^ + 2hxy + hy"^ + 2gx + 2fy + c = 0. . .(1).
Let {x", y") be any other point on the conic.
350 COORDINATE GEOMETRY.
The equation to the straight line joining this point to
(x\ y) is
2/-2/' = |^(^-^') (2).
Since both (ic', 3/') and (a?", 2/") lie on (1), we have
aaj'2 + "Ihxy' + hy'^ + 2^cc' + 2/2/' + c - (3),
and a£c"2 + 2^a;'y ' + hy'"" + 2^£c" + Ify" + c = (4).
Hence, by subtraction, we have
a {x^ - x'") + 2^ {x'y' - xy) + 5 {y"" - 3/"^;
+ 2^(x'-a.") + 2/(2/'-2/'') = (5).
But
2 (o^y - x"y") = {x' + x") {y' - y") + {x' - x") {y' + y"\
so that (5) can be written in the form
{^ - x") [a {x' + x") + h(y'+ y") + 2^]
+ ky' - y") [h {^' + x") + h{y' + y") + 2/] = 0,
y"-y' ^ a{x' + x") + h(y' + y") + 2g
*■ ^' x" -x' k {x' + x") + b{y' + y") + 2/ "
The equation to any secant is therefore
y y~ h{x' + x") + h{y'+y") + 2f^ a:;...^
To obtain the equation to the tangent at {x', y'), we put
x" = x' and y" = y' in this equation, and it becomes
, ax + hy' + <7 . ,,
y-y =- -7-7 — 1^, — > (^ - ^\
^ ^ hx +hy +f^ '
i. e. {ax + liy' + g) x + (hx + by +f)y
= ax'^ + 2hxy' + hy'^ + gx + fy
— — gx —fy — c, by equation (3).
The required equation is therefore
axx' + h (xy + x'y) +l>yy' + g (x + x) + f (y + y')
+ c = (7).
Cor. 1. The equation (7) may be written down, from
the general equation of the second degree, by substituting
XX foi- a^, yy' for 3/^, xy' + xy for 2xy, x + od for 2a?, and
y-vy iovly. (Cf. Art. 152.)
THE GENEKAL CONIC. POLE AND POLAR. 351
Cor. 2. If the conic pass through the origin we have
c = 0, and then the tangent at the origin (where x =0 and
y = 0) is gx +fy = 0,
i. e. the equation to the tangent at the origin is obtained by
equating to zero the terms of the lowest degree in the
equation to the conic.
373. The equation of the previous article may also be obtained
as follows ; If {x', y') and {x", y") be two points on the conic section,
the equation to the line joining them is
a{x-xf){x-x") + hl{x-x'){y-y") + {x-x"){ij-y')] + h{y-y'){y-y")
= ax^ + 2hxy + by^ + 2gx + 2fy + c (1).
For the terms of the second degree on the two sides of (1) cancel,
and the equation reduces to one of the first degree, thus representing
a straight line.
Also, since {x', y') lies on the curve, the equation is satisfied by
putting x=x' and y=y'.
Hence {x', y') is a point lying on (1).
So {x'\ y") lies on (1).
It therefore is the straight line joining them.
Putting x" — x' and y" =y' we have, as the equation to the tangent
at {x', y'),
a{x- x'f + 27i {x - x') [y - if) + & {y - y'f
= ax^ + ^hxy + 'by'^ + 2gx + 2fy + c,
i.e. 2axx' + 2}i {x'y + xy') + 2\)yy' + 2gx + 2fy + c
= ax"^ + 2hx'y' + hy"^
— - 2gx' - 2fy' - c, since {x', y') lies on the conic.
Hence the equation (7) of the last article.
374. To find the condition that any straight line
lx + my + n = (1),
may touch the conic
ax^ + 2hxy + by^ + 2gx + 2fy + c = (2).
Substituting for y in (2) from (1), we have for the equation giving
the abscissae of the points of intersection of (1) and (2),
x^ {am? - 2hlm + hP) - 2x (hmn - bin - gm^ +flm)
+ bn^-2fmn + cm^=0 (3).
If (1) be a tangent, the values of x given by (3) must be equal.
The condition for this is, (Art. 1,)
{hmn - bin - gm^+fhrif={am? - 2hlm + bl^) {bn^ - 2fmn + cm^).
352 COORDINATE GEOMETRY.
On simplifying, we have, after division by iii^,
p {jic -p) + m2 {ca - p2) + n^ [ah - li') + 2mn [gh - af) + 2wZ {lif - hg)
+ 2lm{fg-ch) = 0.
Ex. Find the equations to the tangents to the conic
x^ + 4:xy + 3y^-5x-Gy + S = (1),
ivhich are parallel to the straight line a; + 4i/ = 0.
Tlie equation to any such tangent is
a; + 4^ + c = (2),
where c is to be determined.
This straight line meets (1) in points given by
3x2 _ 2x (5c + 28) + 3c2 + 24c + 48 = 0.
The roots of this equation are equal, i.e. the line (2) is a tangent,
if {2(5c + 28)}2 = 4. 3.(3c2 + 24c+48), i.e. if c=-5 or -8.
The required tangents are therefore
a; + 4?/-5 = 0, and x + iij -Q — 0.
375. As in Arts. 214 and 274 it may be proved that
the polar of {x\ y) with respect to ^ (cc, y) = is
{ax' + hy -{■ g) X + {hx + by' +/) y + gx +/y' + c = 0.
The form of the equation to a polar is therefore the
same as that of a tangent.
Just as in Art. 217 it may now be shewn that, if the
polar of F passes through T, the polar of T passes through
P.
The chord of the conic which is bisected at {x\ y),
being parallel to the polar of (x, y) [Arts. 221 and 280],
has as equation
{ax + hy +g){x- x) + {hx' + by +/){y — y) = 0.
376. To find the equation to the diameter bisecting all
chords parallel to the straight line y = mx.
Any such chord is y = 7nx->rK (1).
This meets the conic section
ax^ + 2hxy + by"^ + 2gx + 2fy + c =
in points whose abscissae are given by
ax^ + ^hx {mx + K) + b {mx + Kf + 2gx + 2f{mx + K)+ g=^0,
i. e. by x^ {a + 2hm + bm^) + 2x {hK + bmK + ^ + fm)
+ blO + 2fK-\-c = 0.
THE GENERAL CONIC. CONJUGATE DIAMETERS. 353
If fl?i and x^ be the roots of this equation, we therefore
have
X ^x - cy{ h + hm)K + g+fm
^ ^ a + 2hm + bm^
Let {X, Y) be the middle point of the required chord,
so that
„ X1 + X2 (h + bm) K+ gf+/7n , .
2 a + 2hm + b7)iF ' ^ ^'
Also, since (X, Y) lies on (1) we have
Y-^mX+K (3).
If between (2) and (3) we eliminate K we have a
relation between X and Y.
This relation is
— (a + 2hm + bm^) X = (h + bm) {Y — mX) + g +fm,
i. e. X [a + hm) + Yih+ bm) + g -¥fm = 0.
The locus of the required middle point is therefore the
straight line whose equation is
x{a + hm) + y(h + bm) + g ■\-fm == 0.
If this be parallel to the straight line y - m'x, we
have
a + Am
m =
•w,
h + bm
i.e. a + h (m + m') + bmm' = (5).
This is therefore the condition that the two straight
lines y — tnx and y — m!x may be parallel to conjugate
diameters of the conic given by the general equation.
377. To find the condition that the pair of straight lines, whose
equation is
Ax'^ + 2Hxy+By^=0 (1),
may be parallel to conjugate diameters of the general conic
ax'^ + 2hxy + by^ + 2gx + 2fy + c = .'. (2).
Let the equations of the straight lines represented by (1) hey = mx
and y = m'x, so that (1) is equivalent to
B [y - mx) [y - m'x) = 0,
and hence m-\-m'= --—, and mm =^ .
L. 23
354 COORDINATE GEOMETRY.
By the condition of the last article it therefore follows that the
lines (1) are parallel to conjugate diameters if
i.e. if Ah-2Hh + Ba = 0.
378. To prove that tioo concentric conic sections always have a
pair, and only one pair, of common conjugate diameters and to find
their equation.
Let the two concentric conic sections be
ax^ + 2hxy + hy'^=l (1),
and a'x^^-2h'xy + l)Y = l (2).
The straight lines
Ax^ + 2Hxy + By^ = (3),
are conjugate diameters of both (1) and (2) if
Ab-2Hh + Ba = 0,
and Ab'-2Hh' + Ba' = 0.
Solving these two equations we have
A -2H__ B
ha' - h'a ~ W^afb ~ hh' - b'h '
Substituting these values in (3), we see that the straight lines
x^ha'-h'a)-xy{ab'-a'b)4-y^{bh'-b'h) = (4)
are always conjugate diameters of both (1) and (2).
Hence there is always a pair of conjugate diameters, real, coinci-
dent, or imaginary, which are common to any two concentric conic
sections.
EXAMPLES. XLII.
1. How many other conditions can a conic section satisfy when
we are given (1) its centre, (2) its focus, (3) its eccentricity, (4) its
axes, (5) a tangent, (6) a tangent and its point of contact, (7) the
position of one of its asymptotes?
2. Find the condition that the straight line lx + niy = l may
touch the parabola {ax-bijf-2 {a^ + &-) (ax + by) + {a^ + b'^f = 0, and
shew that if this straight line meet the axes in P and Q, then PQ
will, when it is a tangent, subtend a right angle at the point {a, b).
3. Two parabolas have a common focus ; prove that the perpen-
dicular from it upon the common tangent passes through the
intersection of the directrices.
[EXS. XLII.] INTERSECTIONS OF TWO CONICS. 355
4. Shew that the conic -^ H r- cos a + i-:-, = sin^ a is inscribed in
a^ ab h^
the rectangle, the equations to whose sides are x^ = a^ and y^ = h^, and
that the quadrilateral formed by joining the points of contact is of
constant perimeter 4 sja^ + 6^, whatever be the value of a.
5. A variable tangent to a conic meets two fixed tangents in two
points, P and Q ; prove that the locus of the middle point of PQ is a
conic which becomes a straight line when the given conic is a parabola.
6. Prove that the chord of contact of tangents, drawn from an
external point to the conic ax^ + 2hxy + hy'^=l, subtends a right angle
at the centre if the point lie on the conic
a;2 (a2 + 7i2) + 2h {a + h) xy+y'^ {h'^ + h^) = a + b.
7. Given the focus and directrix of a conic, prove that the polar
of a given point with respect to it passes through another fixed point.
8. Prove that the locus of the centres of conies which touch the
axes at distances a and b from the origin is the straight line ay = bx.
9. Prove that the locus of the poles of tangents to the conic
ax'^ + 2hocy + by^=l with respect to the conic a'x"^ + 2h'xy + b'y^ = l is
the conic
a {Ji'x + b'yf - 27i [a'x + h'y) {h'x + b'y) + b {a'x + h'yf=db - h\
10. Find the equations to the straight lines which are conjugate
to the coordinate axes with respect to the conic Ax^-{-2Hxy + By^ = l.
Find the condition that they may coincide, and interpret the
result.
11. Find the equation to the common conjugate diameters of the
conies (1) a;2 + 4a;y + 6!/2=l and 2x'^ + &xy + ^y^ — l,
and (2) 2x^-^xy + ^y'^ = l and 2x'^ + ^xy -^y^=l.
12. Prove that the points of intersection of the conies
ax^ ^^hxy + by^=l and a'x^ + 2h'xy-\-b'y'^ = l
are at the ends of conjugate diameters of the first conic, if
ab'-{-a'b-2hh' = 2{ah-U'^).
13. Prove that the equation to the equi-conjugate diameters of
„ ^, , „ ^. ax^ + 2hxy + by^ 2(x^ + y'^)
the conic ax^ + 2kxy + by"^ = 1 is r— f^ — — = — ^ — rr-^ •
^ ^ ab-lv- a + b
379. Two conies, in general, intersect in four points,
real or imaginary.
For the general equation to two conies can be written
in the form
aay^ + 2x Qiy + g) + hy^ + 2fy + c = 0,
and a'x^ + 2x (Ky + g) + 6^' + %f'y + C - 0.
23—2
356 COORDINATE GEOMETRY.
Eliminating x from these equations, we find that the
result is an equation of the fourth degree in y, giving
therefore four values, real or imaginary, for y. Also, by
eliminating x^ from these two equations, we see that there
is only one value of x for each value of y. There are there-
fore only four points of intersection.
380. Equation to any conic passing through the inter-
section of two given conies.
Let S^ ax"" + 2hxy + hy"" + 2gx + ^fy + c = (1),
and S' E ax" + 2h'xy + h'y^ + 2g'x + If'y + c' = . . . (2)
be the equations to the two given conies.
Then ^-_X.S" = (3)
is the equation to any conic passing through the inter-
sections of (1) and (2).
For, since S and aS^' are both of the second degree in x
and 2/, the equation (3) is of the second degree, and hence
represents a conic section.
Also, since (3) is satisfied when both ^S' and S' are zero,
it is satisfied by the points (real or imaginary) which are
common to (1) and (2).
Hence (3) is a conic which passes through the intersec-
tions of (1) and (2).
381. To find the equations to the straight lines passing
through the intersections of two conies given hy the general
equations.
As in the last article, the equation
{a - \a) 0^2 + 2 (A - Xh') xy + (h - \h') ^ + 2 (^ - Xg') x
+ 2(f-\r)y + {c-Xc') = (1),
represents some conic through 4}he intersections of the given
conies.
Now, by Art. 116, (1) represents straight lines if
{a - \a') (b - Kb') (c - Xc') + 2 (f - Xf) (g - Xg') (h - Xh')
-{a- Xa') (/- Xf'f -{b- Xb') (g - Xg'f - (c - Xc') (h - XhJ
= (2).
INTERSECTIONS OF TWO CONIGS. 357
Now (2) is a cubic equation. The thi-ee values of \
found from it will, when substituted successively in (1),
give the three pairs of straight lines which can be drawn
through the (real or imaginary) intersections of the two
conies.
Also, since a cubic equation always has one real root,
one value of A. is real, and it could be shown that there can
always be drawn one pair of real straight lines through the
intersections of two conies.
382. All conies tvhich pass through the intersections of
two rectangular hyperbolas are themselves rectangular hyper-
bolas.
In this case, \i S = and S' — be the two rectangular
hyperbolas, we have
« + 6 = 0, and a +b' = 0. (Art. 358.)
Hence, in the conic S — X;S" — 0, the sum of the co-
efficients of x^ and y"^
= (a- Xa') + {b- Xb') =:(a + b)-X{a' + b') - 0.
Hence, the conic S ~ XS' ^ 0, i.e. any conic through the
intersections of the two rectangular hyperbolas, is itself a
rectangular hyperbola.
Cor. If two rectangular hyperbolas intersect in four points
A, B, G, and D, the two straight lines AD and BC, which are a conic
through the intersection of the two hyperbolas, must be a rectangular
hyperbola. Hence AD and BC must be at right angles. Similarly,
BD and GA, and GD and AB, must be at right angles. Hence D is
the orthocentre of the triangle ABG.
Therefore, if two rectangular hyperbolas intersect in four points,
each point is the orthocentre of the triangle formed by the other
three.
383. 7/* Z = 0, i/== 0, ^= 0, and 11 = be the equations
to the four sides of a quadrilateral taken in order, the
equation to any conic passing through its angular points is
LN=X.MR (1).
For L = passes through one pair of its angular points
and N=0 passes through the other pair. Hence LN = is
the equation to a conic (viz. a pair of straight lines) passing
through the four angular points.
358 COORDINATE GEOMETRY.
Similarly MR - is the equation to another conic
passing through the four points.
Hence LN = X . MR is the equation to any conic through
the four points.
Geometrical meaning. Since L is proportional to the perpen-
dicular from any point {x, y) upon the straight line Z/ = 0, the
relation (1) states that the product of the perpendiculars from any
point of the curve upon the straight lines JL = and N=0 is propor-
tional to the product of the perpendiculars from the same point upon
i!I=0 and E = 0.
Hence If a conic circumscribe a quadrilateral^ the ratio of the
p7'oduct of the perpendiculars from any point P of the conic upon two
opposite sides of the quadrilateral to the product of the perpendiculars
from P upon the other tioo sides is the same for all positions of P.
384. Equations to the conic sections passing through
the intersections of a conic and two
given straight lines.
Let *S' = be the equation to the
given conic.
Let u-0 and v = be the equa-
tions to the two given straight lines
where
u = ax + hy ■\- Cj
and V = ax + Vy -f- c' .
Let the straight line ^^ = meet the conic /S' = in the
points P and P, and let v = meet it in the points Q and T.
The equation to any conic which passes through the
points P, Q^ R, and T will be of the form
S^X.u.v (1).
For (1) is satisfied by the coordinates of any point
which lies both on S — and on u-O] for its coordinates
on being substituted in (1) make both its members zero.
But the points P and R are the only points which lie
both on S — and on u = 0.
The equation (1) therefore denotes a conic passing
through P and R.
Similarly it goes through the intersections of ^S' = and
-y = 0, i. e. through the points Q and T.
THE EQUATION S=XuV. 359
Thus (1) represents some conic going through the four
points P, Q, B, and T.
Also (1) represents any conic going through these four
points. For the quantity A. may be so chosen that it shall
go through any fifth point, or to make it satisfy any fifth
condition* also five conditions completely determine a conic
section.
"Ex.. Find the equation to the conic which passes through the point
(1, 1) and also through the intersections of the conic
tvith the straight lines 2x-y - 5 = and Bx + y -11 = 0. Find also
the paradolas passing through the same points.
The equation to the required conic must by the last article be of
the form
x^ + 2xy + 5y^-7x-8y + Q = X {2x -y - 5) {Sx + y -11) ... {1).
This passes through the point (1, 1) if
l + 2 + 5-7-8 + 6=X{2-l-5) (3 + 1-11), i.e. i{ \=-^\.
The required equation then becomes
28{x^+2xy + 5y^-7x-8y + 6) + {2x-y-5) {3x + y -11) = 0,
i.e. 34x2 + 55xy + l'62if - 233a; - 218y + 223 = 0.
The equation to the required parabola will also be of the form (1),
i.e.
x^{l-Q\)+xy{2 + \)+if{5 + \)-x{7-SU)-y{8 + 6\) + Q-55\ = 0.
This is a parabola (Art. 357) if (2 + X)2 = 4 (1 - 6X) (5 + X),
I.e. if X=|[-12±4V101.
Substituting these values in (1), we have the required equations.
385. Particular cases of the equation
S = Xuv.
I. Let ^t = and v = intersect on the curve, i.e. in
the figure of Art. 384 let the
points P and Q coincide.
The conic S = Xuv then goes
through two coincident points
at P and therefore touches the
original conic at P as in the
figure.
II. Let ^6 = and v =
coincide, so that v = u.
360
COORDINATE GEOMETRY.
In this case the point T also moves up to coincidence
with R and the second conic
touches the original conic at both ^^--''" ~^ \
the points P and B,.
The equation to the second
conic now becomes S=\v?.
When a conic touches a second
conic at each of two points, the
two conies are said to have double
contact with one another.
The two conies S = \v? and /S' = therefore have double
contact with one another, the straight line 16 = passing
through the two points of contact.
As a particular case we see that if ?// — 0, v - 0, and
^f; = be the equations to three straight lines then the
equation vw = \v? represents a conic touching the conic
'DW = where u = meets it, i. e. it is a conic to which
v = and w
contact.
are tangents and .u = is the chord of
tl=rO
III. Let u = be a tangent to the original conic.
In this case the two points P
and P coincide, and the conic
S=-Xuv touches S=0 where u=0
touches it, and v = is the equa-
tion to the straight line joining
the other points of intersection of
the two conies.
If, in addition, v — goes
through the point of contact of w = 0, we have the equation
to a conic which goes through three coincident points at P,
the point of contact of u = ; also the straight line
joining P to the other point of intersection of the two
conies is v = 0.
IV. Finally, let v = and u = coincide and be
tangents at P. The equation S = Xu^ now represents a
conic section passing through four coincident points at the
point where u-0 touches S = 0.
LINE AT INFINITY. 861
386. Line at infinity. We have shewn, in Art.
60, that the straight line, whose equation is
().x + 0.y + G = 0,
is altogether at an infinite distance. This straight line is
called The Line at Infinity. Its equation may for brevity
be written in the form C = 0.
We can shew that parallel lines meet on the line at
infinity.
For the equations to any two parallel straight lines
are
Ax + By + C =0 (1),
and Ax + By+G' = (2).
Now (2) may be written in the form
Ax + By + C + ^ ~^ (0 . a; + . 3/ + (7) = 0,
and hence, by Art. 97, we see that it passes through the
intersection of (1) and the straight line
0.x + 0.y + C = 0.
Hence (1), (2), and the line at infinity meet in a point.
387. Geometrical rtieaning of the equation
S=Xu (1),
where X is a constant^ ayid u = is the equation of a straight
line.
The equation (1) can be written in the form
/S' = Xz* X (0 . £c + . 2/ + 1),
and hence, by Art. 384, represents a conic passing through
the intersection of the conic S—^ with the straight lines
u = ^ and 0.£c + 0.2/ + l = 0.
Hence (1) passes through the intersection of aS'=0 with
the line at infinity.
Since aS' = and S = \u have the same intersections with
362 COORDINATE GEOMETRY.
the line at infinity, it follows that these two conies have
their asjTiiptotes in the same direction.
Particular Case. Let
S = x^ + y^ — a^,
so that S = represents a circle.
Any other circle is
x^ + y^- 2gx — 2 ft/ + c = 0,
i.e. x^ + y^ — a^=2gx+2f'i/ — a^ — c,
so that its equation is of the form *S' = Xil
It therefore follows that any two circles must be looked
upon as intersecting the line at infinity in the same two
(imaginary) points. These imaginary points are called the
Circular Points at Infinity.
388. Geometrical meaning of the equation S = \ where
\ is a constant.
This equation can be written in the form
S^\{0.x + 0.y+lY,
and therefore, by Art. 385, has double contact with S =
where the straight line .x + Q .y +1 =0 meets it, i.e. the
tangents to the two conies at the points where they meet
the line at infinity are the same.
The conies S=0 and S — X therefore have the same
(real or imaginary) asymptotes.
Particular Case. Let S -=^0 denote a circle. Then
S — X (being an equation which differs from S —0 only in
its constant term) represents a concentric circle.
Two concentric circles must therefore be looked upon as
touching one another at the imaginary points where they
meet the Line at Infinity.
Two concentric circles thus have double contact at the
Circular Points at Infinity.
EXAMPLES. 363
EXAMPLES. XLIII.
1. What is the geometrical meaning of the equations S — \. T,
and S = u^ + hu, where ;S = is the equation of a conic, T = is the
equation of a tangent to it, and w = is the equation of any straight
line ?
2. If the major axes of two conies be parallel, prove that the
four points in which they meet are concyclic.
3. Prove that in general two parabolas can be drawn to pass
through the intersections of the conies
ax^ + ^hxy + ly^-^-^gx+^fy+c^Q
and a'x- + 2h'xy + I'y"^ + 2g'x + 2fy + c' = 0,
and that their axes are at right angles if h {a' - V) = h' (a-b).
4. Through the extremities of two focal chords of an ellipse a
conic is described ; if this conic pass through the centre of the ellipse,
prove that it will cut the major axis in another fixed point.
5. Through the extremities of a normal chord of an ellipse a
circle is drawn such that its other common chord passes through the
centre of the ellipse. Prove that the locus of the intersection of
these common chords is an elHpse similar to the given ellipse. If the
eccentricity of the given ellipse be ,^2 (>y2 - 1), prove that the two
ellipses are equal.
6. If two rectangular hyperbolas intersect in four points A, B, G,
and D, prove that the circles described on AB and CD as diameters
cut one another orthogonally.
7. A circle is drawn through the centre of the rectangular
hyperbola xy = c^ to touch the curve and meet it again in two points ;
prove that the locus of the feet of the perpendicular let fall from the
centre upon the common chord is the hyperbola 4^xy = c'^.
8. If a circle touch an ellipse and pass through its centre, prove
that the rectangle contained by the perpendiculars from the centre of
the ellipse upon the common tangent and the common chord is
constant for all points of contact.
9. From a point T whose coordinates are {x', y') a pair of
tangents TP and TQ are drawn to the parabola y'^ = 4:ax; prove that
the liae joining the other pair of points in which the circumcircle of
the triangle TPQ meets the parabola is the polar of the point
(2a -a;', y'), and hence that, if the circle touch the parabola, the line
FQ touches an equal parabola.
10. Prove that the equation to the circle, having double contact
with the ellipse — „ + ^ = 1 at the ends of a latus rectum, is
x^ + i/-2ae^x = a^{l-e^--e^).
SQ"^ COORDINATE GEOMETRY. [EXS. XLIII.]
11. Two circles have double contact "with a conic, their chords of
contact being parallel. Prove that the radical axis of the two circles
is midway between the two chords of contact.
12. If a circle and an ellipse have double contact with one another,
prove that the length of the tangent drawn from any point of the
ellipse to the circle varies as the distance of that point from the
chord of contact.
13. Two conies, A and B, have double contact with a third conic
C. Prove that two of the common chords of A and B, and their
chords of contact with G, meet in a point.
14. Prove that the general equation to the ellipse, having double
contact with the circle x^ + y^=a^ and touching the axis of x at the
origin, is c^x^ + (a- + c^) y^ - 2ahy = 0.
15. A rectangular hyperbola has double contact with a fixed
central conic. If the chord of contact always passes through a fixed
point, prove that the locus of the centre of the hyperbola is a circle
passing through the centre of the fixed conic.
16. A rectangular hyperbola has double contact with a parabola ;
prove that the centre of the hyperbola and the pole of the chord of
contact are equidistant from the directrix of the parabola.
389. To find the equation of the jyair of tangents that
can he drawn from any 'point {x, y') to the general conic
<l> (x, y) = ao(f + 2hxy + hy"^ + 2gx + 2fy + c = 0.
Let T be the given point (x, y'), and let P and R be the
points where the tangents from
T touch the conic. /
The equation to PR is there-
fore ^^ = 0,
where u = (ax' + hy -\-g)x
+ {hx + hy +/) y + gx +fy + c.
The equation to any conic
which touches ^S' = at both of
the points P and R is
S=\u\ (Art. 385),
^. e. ao(? + ^hxy + hy^ + '^gx + Ify + c
— X \{chx + hy + ^) cc -f- {hx +hy' +f)y + gx +fy' + c]^
(!)•
Now the pair of straight lines TP and TR is a conic
I
DIRECTOR CIRCLE. 365
section which touches the given conic at P and R and
which also goes through the point T.
Also we can only draw one conic to go through five
points, viz. T^ two points at P, and two points at R.
If then we find X so that (1) goes through the point T^
it must represent the two tangents TP and TR.
The equation (1) is satisfied by od and y' if
ax"^ + Ihxy' + hy"^ + ^gx + ^fy + c
- X \ax'' + IJixy + hij"' + 2^a^' + 2fy + c]^,
i.e. if ^'=-irr'> — ^\-
The required equation (1) then becomes
^ (x, y) [ax^ + 2hxy + by^ + 2gx + 2fy -f c]
= [(ax + hy' +g)x+ {hx +hy' -\-f)y + gx +fy' + cf,
i.e. ^ (x, y) X ^ (x', y) = u2,
where tt = is the equation to the chord of contact.
390. Director circle of a conic given hy the general
equation of the second degree.
The equation to the two tangents from {x ^ y) to the
conic are, by the last article,
^f? \a^ [x', y') — (ax + hy' + gY]
+ 2xy [hcj) (x, y') - (ax' + hy + g) (hx + by' +/)]
+ 2/' [b4> (x, y) - (hx + by' +fY\ + other terms = 0...(1).
If (x', y') be a point on the director circle of the conic,
the two tangents from it to the conic are at right angles.
Now (1) represents two straight lines at right angles if
the sum of the coefficients of x^ and y^ in it be zero,
i.e. if (a + b)<t> (x',y') - (ax + hy' + gf - (hx' + by' +ff = 0.
Hence the locus of the point (x', y) is
(a + b) (ax^ + 2hxy + by^ + 2gx + 2fy + c)
— (ax + hy + gf — (hx + by +fY — 0,
i.e. the circle whose equation is
(x' + f) (ab - A^) + 2x (bg -fh) + 2y (af- gh)
366 COORDINATE GEOMETRY.
Cor. If the given conic be a parabola, then ah = h^,
and the locus becomes a straight line, viz. the directrix of
the parabola. (Art. 211.)
391. The equation to the director circle may also be obtained in
another manner. For it is a circle, whose centre is at the centre of
the conic, and the square of whose radius is equal to the sum of the
squares of the semi-axes of the conic.
The centre is, Art. 352, the point (%^^ , ^^'H ) .
\ab - Ir ab - h^J
Also, if the equation to the conic be reduced to the form
ax^ + 2hxy + hy^ + c' = 0,
and if a and /3 be its semi-axes, we have, (Art. 364,)
i l-^il^ A 1 _ ah-li'^
^ + ^- _c" ^"^^ ^2--^^2-'
.1.1 T • . o ^o —{a + h) g'
so that, by division, o?-^p^ = — - — ^2 •
The equation to the required circle is therefore
gh-afY_ {a + b)c'
i
{"-w^^'^iy
ah ~ h^J ab - h^
{a + h) (abc + 2fgh - af - bg^ - ch^)
{ab - hP-f
(Art. 352).
392. The equation to the (imaginary) tangents drawn
from the focus of a conic to touch the conic satisfies the
analytical condition for being a circle.
Take the focus of the conic as origin, and let the axis of
X be perpendicular to its directrix, so that the equation to
the latter may be written in the form x-{-h = 0.
The equation to the conic, e being its eccentricity, is
therefore x^ + y^= e^ {x + ky,
i.e. jK2(l-e2) + 2/2-2e'^x-e2P^0.
The equation to the pair of tangents drawn from the
origin is therefore, by Art. 389,
[x^ (1 - e^) + 2^2 _ 2e%x - e^P] [- e'k'''] = [- e^kx - e'^JcJ,
i.e. a^ (1 — e^) + 2/^ — 2e^kx — e^k^ - — e^[x + k]^,
i.e. x' + y^ = (1).
Here the coefficients of x^ and y"^ are equal and the
coefficient of xy is zero.
FOCI OF THE GENERAL CONIC. 367
However the axes and origin of coordinates be changed,
it follows, on making the substitutions of Art. 129, that in
(1) the coefficients of o(? and y"^ will still be equal and the
coefficient of xy zero.
Hence, whatever be the conic and however its equation
may be written, the equation to the tangents from the focus
always satisfies the analytical conditions for being a circle.
393. To find the foci of the conic given hy the general
equation of the second degree
ax^ + ^hxy + hy'^ + "Igx + ^fy + c = 0.
Let (aj', 2/') be a focus. By the last article the equation
to the pair of tangents drawn from'it satisfies the conditions
for being a circle.
The equation to the pair of tangents is
^ (^', y) [«^ + ^hxy + hy'^ + Igx + Ify + c]
= \x {ax' + hy' + g) + y (hx' + by' +/) + (gx' +fy + c)]^.
In this equation the coefficients of xF and y^ must be
equal and the coefficient of xy must be zero.
We therefore have
acf> (x\ y') — (ax + hy' + gY = hcf> (x', y) — {lix + hy +y* )^,
and 7i0 (a?', y'^ = (ax' + hy' + g) (Jix + by' +f),
i.e.
{ax' + hy' + gf - {hx + by' +ff _ {ax' + hy' + g) {hx' + by' +/)
a — b h
= <!>¥,&;) ■••••w-
These equations, on being solved, give the foci.
Cor. Since the directrices are the polars of the foci,
we easily obtain their equations.
394. The equations (4) of the previous article give, in general,
four values for x' and four corresponding values for y'. Two of these
would be found to be real and two imaginary.
In the case of the ellipse the two imaginary foci lie on the minor
axis. That these imaginary foci exist follows from Art. 247, by
writing the standard equation in the form
368 COORDINATE GEOMETRY.
This shews that the imaginary point {0, sj^^-o^] is a focus, the
imaginary line y . =0 is a directrix, and that the correspond-
ing eccentricity is the imaginary quantity
/&2_a2
Similarly for the hyperbola, except that, in this case, the eccen-
tricity is real.
In the ease of the parabola, two of the foci are at infinity and are
imaginary, whilst a third is at infinity and is real.
395. Ex. 1. Find the focus of the parabola
16a;2 - 2ixy + %2 - 80a; - 140?/ -1- 100=0.
The focus is given by the equations
{IQx' - 1 2y' - 40)^ - ( - 12a;^ + 9y' - 70 )^
rj
_ {16x' - 12y ^ - 40) ( - 12a;^ + 9y' - 70)
~ -12
= lQx'^-24:xY + 9y'^-80x'-U0y' + 100 (1).
The first pair of equation (1) give
12 (16a;' - 12y' - 40)2 + 7 (ig^/ _ i2y' _ 40) ( - 12a;' + 9i/ - 70)
-12 (-12a;' -1-9?/' -70)2=0,
i.e. {4 (16a;' - 12y' - 40) - 3 ( - 12a;' + 9y' - 70)}
X { 3 (16a;' - 12^/' - 40) 4- 4 ( - 12ic' -h %' - 70) } = 0,
i.e. (100a;'-75?/' + 50)x(-400) = 0,
so that y = — ^ — .
We then have 16a;' - 12^' - 40 = - 48,
and - 12a;' -[-V- 70 =-64.
The second pair of equation (1) then gives
48 X 64
_ ^ox»^ ^ ^, ^^g^, _ ^2^, _ 40) + 2/' ( - 12a;' + 9?/' - 70) - 40a;' - 70^' + 100
= - 48a;' - 64?/' - 40a;' - 70?/' + 100
= - 88a;' -134?/'-}- 100,
^^n 00 , 536a;' -I- 268 ,^^
i.e. -256= -88a;' ~- -t-100,
o
so that a;' =1, and then 2/' =2.
The focua is therefore the point (1, 2).
AXES OF THE GENERAL CONIC. 369
In the case of a parabola, we may also find the equation to the
directrix, by Art. 390, and then find the coordinates of the focus,
which is the pole of the directrix.
Ex. 2. Find the foci of the conic
55a;2 - SOxy + 39^2 - 40a; - 24^/ - 464=0.
The foci are given by the equation
(55a;^ - 15y' - 20)^ - ( - 15a;^ + S9t/ - 12)2
16
_ {55c(/ - 15y' - 20) ( - 15x' + 39?/^ - 12)
-15
= 55a;'2 - 30a;y + 39?/'2 - 40a;' - 24^/' - 464 (1).
The first pair of equations (1) gives
15 (55a;' - 15^' - 20)2 + ig (55^/ _ 15^' _ 20) ( - 15a;' + 39?/' - 12)
- 15 ( - 15a;' + S9y' - 12)2 ^ q,
i.e. {5 (55a;' - 15y' - 20) - 3 ( - 15a;' + 39^' - 12) }
{3 (55a;' - 15?/' - 20) + 5 ( - 15a;' + 39?/' - 12)} = 0,
i.e. (5a;'-3?/'-l)(3a;' + 5?/'-4) = 0.
, 5a;' - 1
••• ^--3" (2)'
3a;' -4
or y'= — 5-- (3).
Substituting this first value of y in the second pair of equation (1),
we obtain
o/;/o / 1X2 340a;'2- 340a;' -1355
- 25 (2a;' - 1)2= ,
giving a;' = 2 or - 1, Hence from (2) y' — B or - 2.
On substituting the second value of y' in the same pair of equation
(1), we finally have
2a;'2-2a;' + 13 = 0,
the roots of which are imaginary.
We should thus obtain two imaginary foci which would be found
to lie on the minor axis of the conic section. The real foci are
therefore the points (2, 3) and ( - 1, - 2).
396. Equation to the axes of the general
conic.
By Art. 393, the equation
(ax + hy + gf — (hx + by +fy _ {ax ■\-hy-\-g) {hx + hy +/)
a — b h
(1)
represents some conic passing through the foci.
L. 24
370
COORDINATE GEOMETRY.
But, since it could be solved as a quadratic equation to
ax + hy + a . ^ ^ ^ . i , t
give 7 j^ — ^, it represents two straight lines.
ihoc "T" oy ~r J
The equation (1) therefore represents the axes of the
general conic.
397. To find the length of the straight lines drawn
through a given point in a given direction to vieet a given
conic.
Let the equation to the conic be
<^ (aj, y) = ax^ + 2hxy + hy- + 2gx + Ify + c = . . .(1).
Let F be any point (a;', y')^ and through it let there be
drawn a straight line at an angle Q
with the axis of x to meet the
curve in Q and Q'.
The coordinates of any point
on this line distant r from P
are
£c^ + r cos Q and y' + r sin 0.
(Art. 86.)
If this point be on (1), we
have
a{x' + r cos &f + 2A {x + r cos 0) [y' + r sin 0) + b (y' + r sin 0)^
+ 2g {x + r cos 0) + 2/ {y + r sin ^) + c = 0,
i.e.
r^ [a cos2 e + 2h cos ^ sin 6* + & sin^ 0]
+ 2r [(ax + hy' + g) cos d + {hx + by +/) sin ^] + <^ (a;', 3/') =
^ ;-(2).
For any given value of this is a quadratic equation in
r, and therefore for any straight line drawn at an inclina-
tion it gives the values of FQ and PQ'.
If the two values of r given by equation (2) be of
opposite sign, the points Q and Q' lie on opposite sides
of P.
If P be on the curve, then <f) (x, y) is zero and one value
of T obtained from (2) is zero.
Y
R
l>rr
p
C
) X
INTERCEPTS ON LINES. 371
398. If two chords PQQ' and PRR' he drawn in given
directions through any pohit P to meet the curve in Q, Q' and
R^ R' respectively, the ratio of the rectangle PQ . PQ' to the
rectangle PR . PR' is the same for all points, o^nd is therefore
equal to the ratio of the squares of the diameters of the conic
which are drawn in the given directions.
The values of PQ and PQ' are given by the equation of
the last article, and therefore
PQ . PQ' = product of the roots
^ ^ {^\ y) /jx
a cos^ 6 + 2h cos sin ^ + 6 sin^ 6'"^ ''
So, if PRR' be drawn at an angle & to the axis, we have
PR pp'= 9 {^^ y) /o\
a cos^ 0' + 2A cos 0' sin 6' -\-h sin^ 0'"'^ ^'
On dividing (1) by (2), we have
PQ . PQ' _ a cos^ 6' + 2h cos 0' sin 0' + 1} sin^ 6'
PR . PR' ~ a cos^ e-\-2h cos ^ sin ^ + 6 sin^ 6 '
The right-hand member of this equation does not contain
x' or y', i. e. it does not depend on the position of P but only
on the directions and $\
The quantity ^ ' p> is therefore the same for all
positions of P.
In the particular case when P is at the centre of the
CO'"^
conic this ratio becomes ,^ ■, , where C is the centre and CQ'
OR
and CR" are parallel to the two given directions.
Cor. If Q and Q' coincide, and also R and R', the two
lines PQQ' and PRR become the tangents from P, and the
above relation then gives
PQ^C(r' . PQ _ CQ"
PR' ~ OR'"' ' ""• ^' PR ~ CR" ■
Hence, If two tangents be drawn from a point to a conic,
their lengths are to one another in the ratio of the parallel
semi-diameters of the conic.
24—2
372
COORDINATE GEOMETRY.
399. If PQQ' and P^QiQi ^^ ^'^^ chords drawn in
parcdlel directions from, two points P aiid F^ to meet a conic
in Q and Q', and Q^ and Qi, resjyectively, then the ratio of
the rectangles PQ . PQ' and PiQi . PiQi is independent of the
direction of the chords.
For, if P and P^ be respectively the points (x', y') and
{x", y")y and 6 be the angle that each chord makes with
the axis, we have, as in the last article,
^ (^', y)
PQ.PQ'=^
and P^Q,,P^Q;^
a cos2 e-\-2h cos 6' sin (9 + 6 sin^ '
<i> K, y")
so that
a cos^ 6 -\-'2ih cos 6 sin ^ + 6 sin^ '
PQ.PQ' <i>{x,y')
PiQi-P^Qi ^{x",y"y
400. If a circle and a conic section cut one another in four 'points,
the straight line joining one pair of points of intersection and the
straight line joining the other pair are equally inclined to the axis of
the conic.
For (Fig. Art. 397) let the circle and conic intersect in the four
points Q, Q' and B, B' and let QQ' and RB' meet in P.
But, since Q, Q', B, and B' are four points on a circle, we have
PQ . PQ' = PB . PB'. [Euc. III. 36, Cor.]
.-. CQ" = CB".
Also in any conic equal radii from the centre are equally inclined
to the axis of the conic.
Hence GQ" and CB", and therefore PQQ' and PBB\ are equally
inclined to the axis of the conic.
401. To shew that any chord of a conic is cut har-
monically by the curve, any point on
the chord^ and the pjolar of this point
with respect to the conic.
Take the point as origin, and let
the equation to the conic be
ao(? + ^hxy + hy"^ + 2gx + 2/3/ + c -
(!)>
HARMONIC PROPERTY OF THE POLAR. 373
or, in polar coordinates,
r^ {a cos^ + 2h cos ^ sin ^ + 6 sin^ 6) + 2r {g cos 6 +/sin 6) + c^Q,
i.e.
c. -^+2 .-. (gcosO +/sin 0)
+ a cos^ + 2h cos sin + b sin^O=^ 0.
Hence, if the chord OFF' be drawn at an angle to OX,
we have
TT^i + yc-T^, — sum of the roots of this equation in -
OF OF r
g cos 6 +y sin 6
c
Let -ff be a point on this chord such that
2___1_ J
OF ~ of"" OF' '
Then, if OR = p, we have
2 ^cos^+ysin^
P ^
so that the locus of F is
g . p cos +/. p sin + c — Oj
or, in Cartesian coordinates,
gx+/y + G^O (2).
But (2) is the polar of the origin with respect to the
conic (1), so that the locus of F is the polar of 0.
The straight line FF' is therefore cut harmonically by
and the point in which it cuts the polar of 0.
Ex. Through any point is draion a straight line to cut a conic
in P and P' and on it is taken a point R such that OR is (1) the
arithmetic mean, and (2) the geometric mean, betioeen OP and OP'.
Find in each case the locus of R.
Using the same notation as in the last article, we have
0P+0P'=-2 - g COB d+f Bind
and OP.OP'=
a cos2 + 2h cos ^ sin ^ + & sin^ d *
c
a cos^ 6 + 2h cos 6 sin ^ + & sin^ '
374 COORDINATE GEOMETRY.
(1) If R be the point (p, 6) we have
-i(np,oP'\- g COB d+f sine
p-^[^ujr-tLjjr ) acos'^d + 2hcoseBmd + bs'm^d'
i.e. ap cos2 d + 27ip cos 6 Bin 6 + hp sin^ d + gQOBd-\-f sin ^ = 0,
i.e., in Cartesian coordinates,
ax^ + 2 7ia;i/ + by^+gx +fy = 0.
The locus is therefore a conic passing through and the inter-
section of the conic and the polar of 0, i.e. through the points T
and T', and having its asymptotes parallel to those of the given
conic.
(2) If R be the point {p, 6), we have in this case
/•
„2 — np OP'—
^ ^^.^ acos2^ + 27icos^sin0 + &sin2 6''
i.e. ap^ cos^ d + 2hp^ cos ^ sin ^ + hp- sin^ ^ = c,
i.e. aa:2 + 2/ia;^ + 6?/2 = c.
The locus is therefore a conic, having its centre at and passing
through T and T', and having its asymptotes parallel to those of the
given conic.
402. To find the locus of the middle points of 'parallel cliords of a
conic. [Cf. Art. 376.]
The lengths of the segments of the chord drawn through the point
{x\ y') at an angle 6 to the axis of x is given by equation (2) of Art.
397.
If {x', y') be the middle point of the chord the roots of this
equation are equal in magnitude but opposite in sign, so that their
algebraic sum is zero.
The coefi&cient of r in this equation is therefore zero, so that
{ax' + hy' + g) cos + [hx' + by' +/) sin ^ = 0.
The locus of the middle point of chords inclined at an angle 6 to
the axis of x is therefore the straight line
(ax + hy + g) + {hx + by +f) tan ^ = 0.
Hence the locus of the middle points of chords parallel to the line
y = mx is
{ax + hy +g) + {hx + by+f) m = 0,
i.e. x{a + hm) + {h + bm) y + g +fm = 0.
This is parallel to the line y = m'x if
, a + hm
m = -J- — ^ ,
h + bm
i.e. if a + h{m + m') + bmm' = 0.
This is therefore the condition that y = mx and y — m'x should be
parallel to conjugate diameters.
EXAMPLES. 375
403. Equation to the pair of tangents drawn from a given point
{x', y') to a given conic, [Cf. Art. 389.]
If a straight line be drawn through [x', y'), the point P, to meet
the conic in Q and Q\ the lengths of PQ and PQ' are given by the
equation
r2 (a cos^ ^ + 2/4 cos ^ sin ^ + 6 sin^ 6)
+ 2r l{ax' + hy' + g) cos 6 + {hx' + 5?/' +/) sin ^] + (x', ?/') = 0.
The roots of this equation are equal, i.e. the corresponding Hnes
touch the conic, if
{a cos2 ^ + 2/i cos ^ sin ^ + & sin^ ^) x <^ {x', y')
= [{ax' + hy' + £f) cos 6 + {hx' + by'+f) sin ^p,
I.e. if (a + 27i tan ^ + & tan^ ^) x (a;', y')
= [{ax' + hy' + g) + {hx' + by'+f) tan ef...{l).
The roots of this equation give the corresponding directions of the
tangents through P.
Also the equation to the line through P inclined at an angle 6 to
the axis of x is
^^ = tan0 (2).
x-x
If we substitute for tan 6 in (1) from (2) we shall get the equation
to the pair of tangents from P.
On substitution we have
{aix-x')^ + 2h{x-x'){y-y') + b{y-y'y~}<t>{x',y')
= l{ax' + hy' + g){x- x') + {hx' + by' +/) {y - y')f.
This equation reduces to the form of Art. 389.
EXAMPLES. XLIV.
1. Two tangents are drawn to an ellipse from a point P; if the
points in which these tangents meet the axes of the ellipse be
concyclic, prove that the locus of P is a rectangular hj^erbola.
2. A pair of tangents to the conic Ax^ + By^=l intercept a
constant distance 2 A; on the axis of x ; prove that the locus of their
point of intersection is the curve
By^{Ax^ + By^-l) = Ak^{By^-lf.
3. Pairs of tangents are drawn to the conic ax^ + ^y^=l so as to
be always parallel to conjugate diameters of the conic
ax^ + hxy + by^ = l;
shew that the locus of their point of intersection is the conic
ax" + 2hxy + by^= - + 7; .
a p
376 COORDINATE GEOMETRY. [Exs.
4. Prove that the director circles of all conies which touch two
given straight lines at given points have a common radical axis.
5. A parabola circumscribes a right-angled triangle. Taking its
sides as the axes of coordinates, prove that the locus of the foot of the
perpendicular from the right angle upon the directrix is the curve
whose equation is
2xy {x^ + 2/2) [hy + kx) + hhj^ + A;V= 0,
and that the axis is one of the family of straight lines
m^h - k
y = mx-— K ,
where m is an arbitrary parameter and 2h and 2k are the sides of the
triangle.
Find the foci of the curves
6. 300a;2 + d20xy + 144 </2 - 1220a; - 768?/ + 199 =0.
7. l&x^-2Axy + 9y^ + 28x+14y + 21 = 0.
8. 144.C2 _ i20xy + 25y^ + Qlx - A2y + 13 = 0.
9. x^- &xy + y^- 10a; - lOy -19 = and also its directrices.
10. Prove that the foci of the conic
ax'^ + 2hxy + hy^ = 1
are given by the equations
a-h ~ h ~~ a^-b^'
11. Prove that the locus of the foci of all conies which touch the
four lines x= ^a and y=±b is the hyperbola x^-y^=a^- h^.
12. Given the centre of a conic and two tangents ; prove that the
locus of the foci is a hyperbola.
[Take the two tangents as axes, their inclination being w; let
(^ij Vi) ^^^ {^2> 2/2) ^^ *^^ ^^^^> ^^^ (^' ^) *^® given centre. Then
x-^ + X2=2h and 2/1 + 1/2 = 2^; also, by Art. 270 {j3), we have
y-^y^ sin^ w = x-^x^ sin^ w = (semi-minor axis)^.
From these equations, eliminating x.2 and 2/2, we have
^1^ - y^=2hx^ - 2ky^ ,]
13. A given ellipse, of semi-axes a and h, slides between two
perpendicular lines ; prove that the locus of its focus is the curve
(a;2 + ^2) ^^^2^2 _j. 54) ^ 4aPx^y\
14. Conies are drawn touching both the axes, supposed oblique, at
the same given distance a from the origin. Prove that the foci lie
either on the straight line x = y, or on the circle
x^ + y'^ + 2xycos o}=a{x + y).
15. Find the locus of the foci of conies which have a common point
and a common director circle.
XLIV.] TANGENT AND NORMAL AS AXES. 377
16. Find the locus of the focus of a rectangular hyperbola a
diameter of which is given in magnitude and position.
17. Through a fixed point chords POP' and QOQ' are drawn at
right angles to one another to meet a given conic in P, P' , Q, and Q'.
Prove that ^^, + QO^' ^' ^on^i^ni.
18. A point is taken on the major axis of an ellipse whose abscissa
is ae ~- fj2 — e^ ; prove that the sum of the squares of the reciprocals
of the segments of any chord through it is constant.
19. Through a fixed point is drawn a line OPP' to meet a conic
in P and P' ; prove that the locus of a point Q on OPP', such that
7-rs = Tvvio + y^T^ is another conic whose centre is O.
OQ^ OP^ OP^
20. Prove Carnot's theorem, viz. : If a conic section cut the side
BC of a triangle ABC in the points A' and A'\ and, similarly, the
side CA in B' and B", and AB in C and C'% then
BA' . BA" . GB' . GB" .AG'. AG"=GA' . GA" . AB' . AB" . BG' . BG".
[Use Art. 398.]
21. Obtain the equations giving the foci of the general conic by
making use of the fact that, if >Sf be a focus and PSP' any chord of
the conic passing through it, then -^^^ + — ^, is the same for all direc-
oP hP
tions of the chord.
22- Obtain the equations for the foci also from the fact that the
product of the perpendiculars drawn from them upon any tangent is
the same for all tangents.
404. To find the equation to a conic, the axes of co-
ordinates being a tangent and normal to the conic.
Since the origin is on the curve, the equation to the
curve must be satisfied by the coordinates (0, 0) so that the
equation has no constant term and therefore is of the form
ax^ + 2hxy + h'lf + 2gx + yy — 0.
If this curve touch the axis of x at the orio-in, then,
when 2/ = 0, we must have a perfect square and therefore
The required equation is therefore
ax^ + Ihxy + hy"^ + 2/^/ = (1).
Ex. O is any point on a conic and PQ a chord ; prove that
(1) if PQ subtend a right angle at O, it passes through a fixed
point on the normal at O, and
378
COORDINATE GEOMETRY.
(2) if OP and OQ be equally inclined to the normal at 0, then
PQ passes through a fixed point on the tangent at 0.
Take the tangent and normal at as axes, so that the equation to
the conic is (1).
Let the equation to PQ be y = mx + c (2).
Then, by Art. 122, the equation to the lines OP and OQ is
c {ax'^ + 2hxy + by^) + 2fy {y -mx) = (3).
(1) If the lines OP and OQ be at right angles then (Art. 66), we
have ac + bc + 2/= 0,
I.e.
a + b
= a constant for all positions of PQ.
But c is the intercept of PQ on the axis of y, i.e. on the normal
at 0.
The straight line PQ therefore passes through a fixed point on the
— 2/
normal at which is distant v from O.
a + b
This point is often called the Fregier Point.
(2) If again OP and OQ be equally inclined to the axis of y then,
in equation (3), the coefficient of xy must be zero, and hence
2;ic-2/m = 0,
^ I-
m h
- c
I.e.
= ^ = constant.
But — is the intercept on the axis of x of the line PQ.
m
Hence, in this case, PQ passes through a fixed point on the tangent
at 0.
405. General equation to conies 2^<^ssing through four
given points.
Let A, B, C, and D be the four points, and let BA and
CD meet in 0. Take OAB
and ODC as the axes, and
let OA = \,0B = V, OD - ix,
B.ndOC^fx\
Let any conic passing
through the four points be
ax^i-2h'xi/ + by^
+ 2gx+2/y+c=0...{l).
If we put y = in this
equation the roots of the
resulting equation must be A and A,'.
comes THROUGH FOUR POINTS. 379
Hence 2g = — a (X + X') and c — aX\\
Similarly h — —, , and %f— — c - — y- .
On substituting in (1) we have
fijULof + 2hxi/ + XX'y^ — fxfjL (X + X') x
-XX' {fx + ix)y + XX'fifx' = (1),
where h = h' ^-^ .
G
This is the required equation, h being a constant as yet
undetermined and depending on which of the conies through
A, B, C, and D we are considering.
406. Aliter. We have proved in Art. 383 that the
equation hLN=MR^ ^ being any constant, represents any
conic circumscribing the quadrilateral formed by the four
straight lines L = 0, 31=0, iV= 0, and E = taken in this
order.
With the notation of the previous article the equations
to the four lines AB, BC, CD, and DA are
2/=0, l+y-,-l=0, x = 0,
X fx
and %^_1 = 0.
A fJi
The equation to any conic circumscribing the quadri-
lateral A BCD is therefore
^--K^^O(^^o w,
i.e.
/xfxx"^ + xy (X/jl' + X'/ji - hXX'ixii) + XX' y"^
— n-ix! (X + X') x~ XX' (fji + fx) y + XX'fjbfji = 0.
On putting Xfx + X'jx - kXX'ixfji' equal to another constant
2h we have the equation (1) of the previous article.
380 COORDINATE GEOMETRY.
407. Only one conic can he drawn through any Jive
points.
For the general equation to a conic through four points
is (1) of Art. 405.
If we wish it to pass through a fifth point, we substitute
the coordinates of this fifth point in this equation, and thus
obtain the corresponding value of h. Except when three of
the five points lie on a straight line a value of h will always
be found, and only one.
Ex. Find the equation to the conic section which passes through
the five points A, B, C, D, and E, whose coordinates are (1, 2), (3, —4),
(-1, 3), (-2, -S),and{5, 6).
The equations to AB, BC, CD, and DA are easily found to be
y + 3x-5 = 0, 4:y + 7x-5 = 0, 6x-y + 9 = 0, and 5x-Sy + l = 0.
The equation to any conic through the four points A, B, C, and D
is therefore
{y + Sx-5){6x-y + 9) = \{4:y + 7x-5){5x-3y + l) (1).
If this conic pass through the point E, the equation (1) must be
satisfied by the values x = 5 and y = &.
We thus have X = V- ^^^i on substitution in (1), the required
equation is
223a;2 - 38xy - 123?/ - 171a; + 8Sy + 350 = 0,
which represents a hyperbola.
4:03. To find the general equation to a conic section
which touches four given straight lines, i.e. which is inscribed
in a given quadrilateral.
THE CONIC LM = R~. 381
Let the four straight lines form the sides of the quadri-
lateral ABCD. Let BA and CD meet in 0, and take OAB
and ODC as the axes of x and y, and let the equations to
the other two sides BG and DA be
l-^x + 7n^y -1=0, and l^x + m^y -1 = 0.
Let the equation to the straight line joining the points
of contact of any conic touching the axes at P and Q be
ax + hy —1 = 0.
By Art. 385, II, the equation to the conic is then
2Xxy = (ax + hy — Vf (1).
The condition that the straight line BG should touch
this conic is, as in Art. 374, found to be
A. = 2(a-Zi)(6-mi) (2).
Similarly, it will be touched by AD if
X = 2{a-l){h-m^) (3).
The required conic has therefore (1) as its equation, the
values of a and h being given in terms of the quantity A. by
means of (2) and (3).
Also X is any quantity we may choose. Hence we have
the system of conies touching the four given lines.
If we solve (2) and (3), we obtain
2& - (mi + W2 ) _ _ 2a-(Zi + y _ ^ / -^ 2\
m^ -7112 ~~ h~h ~ ^ (^1 ~ ^2) ("*i ~ ^^2) *
409. The conic LM=R", where L = 0, M=0, and
R — are the equations of straight lines.
The equation LM=0 represents a conic, viz. two straight
lines.
Hence, by Art. 385, II, the equation
LM=R' (1),
represents a conic touching the straight lines Z = 0, and
M = 0, where R = meets them.
382 COORDINATE GEOMETRY.
Thus L — and M=0 are a pair of tangents and jS =
the corresponding chord of contact.
Every point which satisfies the equations M— jx^L and
B — [xL clearly lies on (1).
Hence the point of intersection of the straight lines
M— fjL^L and E — fxL lies on the conic (1) for all values of
fx. This point may be called the point "ju.."
410. To find the equation to the straight line joining
two jyoints "/x" and "/x"' and the equation to the tangent at
the ^point "/x."
Consider the equation
aL + hM-\-R^O (1).
Since it is of the first degree and contains two constants
a and 6, at our disposal, it can be made to represent any
straight line.
If it pass through the point " /x " it must be satisfied by
the substitutions M = fxL and B — fxL.
Hence a + 6/x^ + /x = (2).
Similarly, if it pass through the point " /x' " we have
a + bfx'' + fM' = (3).
Solving (2) and (3), we have
a , — 1
/X/X fX + fX
On substitution in (1), the equation to the joining line is
Lfxfx' + M-(ix + [x')JR = 0.
By putting fx' = fx we have, as the equation to the
tangent at the point "/x,"
Lfx^ + M- 2fxE = 0,
EXAMPLES. 383
EXAMPLES. XLV.
1. Prove that the locus of the foot of the perpendicular let fall
from the origin upon tangents to the conic ax^ + 2hxy + by' = 2x is the
curve {h^ - ah) {x^ + y'^f + 2 {x^ + ^2) {p^ _ /j^) +^f = Q,
2. In the conic ax'^ + 2hxij + by^ = 2y, prove that the rectangle
contained by the focal distances of the origin is — — - „ .
ao — h^
3. Tangents are drawn to the conic ax^ + 2hxy + by'^ = 2x from
two points on the axis of x equidistant from the origin; prove that
their four points of intersection lie on the conic hy^ + hxy=x.
If the tangents be drawn from two points on the axis of y equi-
distant from the origin, prove that the points of intersection are on a
straight Hne.
4. A system of conies is drawn to pass through four fixed points;
prove that
(1) the polars of a given point all pass through a fixed point,
and (2) the locus of the pole of a given line is a conic section.
5. Find the equation to the conic passing through the origin and
the points (1, 1), ( -1, 1), (2, 0), and (3, -2). Determine its species.
6. Prove that the locus of the centre of all conies circumscribing
the quadrilateral formed by the straight lines ^ = 0, x = 0, x-{-y = l,
and 2/ - a;= 2 is the conic 2x^ - 2y^ + Axy + 5?/ - 2 = 0.
7. Prove that the locus of the centres of all conies, which pass
through the centres of the inscribed and escribed circles of a triangle,
is the circumscribing circle of the triangle.
8. Prove that the locus of the extremities of the principal axes of
all conies, which can be described through the four points ( =t a, 0) and
(0, ± h), is the curve
($.-t)(''^^y')=^'-y-
9. A, B, C, and D are four fixed points and AB and CD meet in
O ; any straight line passing through O meets AD and BC in R and
jR' respectively, and any conic passing through the four given points
in S and S' ; prove that
J^ J^_ J^ , 1
OR "^ OR' ~ OS ''" OS' '
10. Prove that, in general, two parabolas can be drawn through
four points, and that either two, or none, can be drawn.
[For a parabola we have 7i= ± ^JXK'fxfx'.]
384 COORDINATE GEOMETRY. [ExS. XLV.
11. Prove that the locus of the centres of the conies ch'cumserib-
ing a quadrilateral ABGD (Fig. Art. 405) is a conic passing through
the vertices 0, L, and M of the quadrilateral and through the middle
points of AB, AC, AD, BG, BD, and CD.
Prove also that its asymptotes are parallel to the axes of the
parabolas through the four points.
[The required locus is obtained by eliminating h from the equa-
tions 2fjufjL'x + 2hy - fifx,' {\ + \')=0, and 2hx + 2X\'y -XV {fjt. + fM') = 0.]
12. By taking the case when XX'= -/">«•' and when AB and CD
are perpendicular (in which case ABC is a triangle having D as its
orthocentre and AL, BM, and CO are the perpendiculars on its
sides), prove that all conies passing through the vertices of a triangle
and its orthocentre are rectangular hyperbolas.
From Ex, 11 prove also that the locus of its centre is the nine
point circle of the triangle.
13. Prove that the triangle OML (Fig. Art. 405) is such that each
angular point is the pole of the opposite side with respect to any
conic passing through the angular points A, B, C, and D of the
quadrilateral.
[Such a triangle is called a Self Conjugate Triangle.]
14. Prove that only one rectangular hjrperbola can be drawn
through four given points. Prove also that the nine point circles of
the four triangles that can be formed by four given points meet in a
point, viz. , the centre of the rectangular hyperbola passing through
the four points.
15. By using the result of Art. 374, prove that in general, two
conies can be drawn through four points to touch a given straight
line.
A system of conies is inscribed in the same quadrilateral ; prove
that
16. the locus of the pole of a given straight line with respect to
this system is a straight line.
17. the locus of their centres is a straight line passing through the
middle points of the diagonals of the quadrilateral.
18. Prove that the triangle formed by the three diagonals OL,
AC, and BD (Fig. Art. 408) is such that each of its angular points is
the pole of the opposite side with respect to any conic inscribed in the
quadrilateral.
19. Prove that only one parabola can be drawn to touch any four
given lines.
Hence prove that, if the four triangles that can be made by four
lines be drawn, the orthocentres of these four straight lines lie on a
straight line, and their circumcircles meet in a point.
CHAPTER XYII.
MISCELLANEOUS PROPOSITIONS.
On the four normals that can be drawn from any point in
the plane of a central conic to the conic.
411, Let the equation to the conic be
Ax' + By^^l (1).
[If A and JB be both positive, it is an ellipse ; if one be
positive and the other negative, it is a hyperbola.]
The equation to the normal at any point (x, y') of the
curve is
X — X y — y'
If this normal pass through the given point (/«,, k), we
have
h — x k — y'
i.e. (A-£)x'y' + £hy'-Akx'=:0 (2).
This is an equation to determine the point (cc', y') such
that the normal at it goes through the point (h, k). It
shews that the point (x, y') lies on the rectangular hyper-
bola
{A-JB)xy + £hy-Akx = (3).
The point (x', y') is therefore both on the curve (3) and
on the curve (1). Also these two conies intersect in four
points, real or imaginary. There are therefore four points,
L. 25
386 COORDINATE GEOMETRY.
in general, lying on (1), such that the normals at them pass
through the given point (A, k).
Also the hyperbola (3) passes through the origin and
the point (A, k) and its asymptotes are parallel to the axes.
Hence From a given point four normals can in general
he drawn to a given central conic, and their feet all lie on a
certain rectangular hyperbola, which jjasses through the
given point and the centre of the conic, and has its asym^ptotes
parallel to the axes of the given conic.
412. To find the conditions that the normals at the
points where two given straight lines meet a central conic
may m,eet in a p)oint.
Let the conic be
A^-\-By''^\ (1),
and let the normals to it at the points where it is met by
the straight lines
l^ + m,^ — \ (2),
and l^x + m^y = 1 (3)
meet in the point (h, k).
By Art. 384, the equation to any conic passing through
the intersection of (1) with (2) and (3) is
Aa^ + By^ - 1 + X {l-^x + m^y — 1) (^2^ + m.^y - 1) = 0...(4).
Since these intersections are the feet of the four
normals drawn from (A, k), then, by the last article, the
conic
(A- £) xy + Bhy - Akx = (5)
passes through the same four points.
For some value of X it therefore follows that (4) and (5)
are the same.
Comparing these equations, we have, since the co-
efficients of a^ and y^ and the constant term in (5) are all
zero,
A + Xl^l^ = 0, B + Xm^m^ ~ 0, and - 1 +X = 0.
Therefore X =^ 1, and hence
lih — ~^i and m^m^ — ~B (6).
CONCURRENT NORMALS. 387
The relations (6) are the required conditions.
Also, comparing the remaining coefficients in (4) and (5),
we have
A-B -Ak " Bh '
7 A—B m-, + mo ,^.
so that h = ^— y-^ — j-^ (7),
B L^m^ + t^^frix
and ],-_.^lz1 h±^ (8).
Cor. 1. If the given conic be an ellipse, we have
-4 = -5 and B = j^.
The relations (6) then give
cH^^ = IP'm^^ = — 1 (9),
and the coordinates of the point of concurrence are
j^^ a^-h'' m^ + m, ^^ 1 - 5^??^l^
and k = 7^ — -^ j - m-^ (a^ - O'') . —j- — 75 — ^.
Cor. 2. If the equations to the straight lines be given
in the form y = m^c + c and 3/ = m'x + c', we have
m — ^, c= — , m = , and c — — .
mi 7/^1 mg ma
The relations (9) then give
77iy/i' = — „ and cc' = — 6^.
413. If f/ie normals at four points P, Q, R, and S of an ellipse
meet in a point, the sum of their eccentric angles is equal to an odd
multiple of two Hght angles.
25—2
388 COORDINATE GEOMETRY.
If a, /3, 7, and 5 be the eccentric angles of the four points, the
equations to JPQ and BS are
cos
.y=.-a..-cot^-+— — — .
sm-^
& cos '-^
and 2/ = - a^ • - cot ^- + i- . [Art. 259.]
a 2 . 7+0
sm^-
Since the normals at these points meet in a point, we have, by
Art. 412, Cor. 2,
_=mm'=-5 cot —^ cot ^hr--
a^ a^ 2 2
. a + iS 7 + 5 ,
*. tan — ^ = cot = tan
a a
(tt_ _ 7+l5\
V2 "27
a+/3 TT 7 + 5
,. __=n7r+---— ,
*.e. a + /3 + 7 + 5=(2w + l)7r.
414. XSx. 1. I/' i/ie normals at the points A, B, C, and D of an
ellipse meet in a point O, prove that SA . SB . SO. SD = \^ . SO^y where
S is one of the foci and X is a constant.
Let the equation to the ellipse be
^i+f:=i (1),
a^ 0'^
and let be the point {h, k).
As in Art. 411, the feet of the normals drawn from lie on the
hyperbola
\a^ b^J
hy kx -
i.e. a^e^xy = aViy — b^kx (2).
The coordinates of the points A, B, G, and D are therefore found
by solving (1) and (2),
From (2) we have y = -o-n 5-\ •
^ ' ^ a^{h-e^x)
Substituting in (1) and simplifying, we obtain
a;4a2g4 _ 2ha^e^x^ + x^ {a^h^ + 62A;2 _ a*e^) + 2;ie2^4^ _ ^4/^2 _ o. . . (3).
CONCURRENT NORMALS. EXAMPLES. 389
If a?!, x^, x^, and x^ be the roots of this equation, we have (Art. 2),
_ 2ha^ _ a^h^
If S be the point ( - ae, 0) we have, by Art. 251,
SA=.a + ex-^.
:. SA.SB.SG.SD = {a + eXj){a+ex.2){a + ex^){a + ex^) '
= a^ + a^Sa?! + a^e^^x^x^ + ae^'Lx-^x^.^ + e'^x^x^^^^
= -^ {(^ + ae)2 + /c2}-, on substitution and simplification,
= k.SOK
Aliter. If p stand for one of the quantities *S^^, SB, SG, or SD
we have p=a + ex,
i.e. x=-{p-a).
Substituting this value in (3) we obtain an equation in the fourth
degree, and easily have
PiPzPsPi^ -2[{h + ^^)^ + ^% ^3 before.
Ex. 2. If the normals at four points P, Q, B, and S of a central
conic meet in a point, and if PQ pass through a fixed point, find the
lotus of the middle point of BS.
Let the equation to PQ be
y = m-^x + Cj^ (1),
and that to BS y = m^-\-C2 ■ (2).
If the equation to the given conic be Ax'^ + By^=l, we then have
(by Art. 412, Cor. 2)
mim2=-g (3),
1^
B
If (/, g) be the fixed point through which PQ passes, we have
g=m^f+Cj^ (5).
Now the middle point of BS lies on the diameter conjugate to it,
i.e. by Art. 376, on the diameter
A
and c-^c^=. - — (4).
Bm,
i.e., by (3), y--m^x (6).
890 COORDINATE GEOMETRY.
Now, from (4) and (5),
c„= -
B{g-M)'
so that, by (3), the equation to BS is
y=mr,''-B{g-fm,) - ^'^^-
Eliminating m^ between (6) and (7), we easily have, as the equation
to the required locus,
{Ax^ + By^){gx+fy)+xy = 0.
Cor. From equation (6) it follows that the diameter conjugate to
ES is equally inclined with PQ to the axis, and hence that the points
P and Q and the ends of the diameter conjugate to BS are coney clic
(Art. 400),
EXAMPLES. XLVI.
1. If the sum of the squares of the four normals drawn from a
point to an ellipse be constant, prove that the locus of is a conic.
2. If the sum of the reciprocals of the distances from a focus of
the feet of the four normals drawn from a point to an ellipse be
4
^-r r > prove that the locus of is a parabola passing through that
focus.
3. If four normals be drawn from a point to an ellipse and if
the sum of the squares of the reciprocals of perpendiculars from the
centre upon the tangents drawn at their feet be constant, prove that
the locus of is a hyperbola.
4. The normals at four points of an ellipse are concurrent and
they meet the major axis in Gj, G^, G^a, and G^; prove that
C(?i "*" CG^ "^ GG^'^CG^~ GG^+GG^+GG^+GG^'
5. If the normals to a central conic at four points L, M, N, and
P be concurrent, and if the circle through L, 31, and N meet the curve
again in P', prove that PP' is a diameter.
6. Shew that the locus of the foci of the rectangular hyperbolas
which pass through the four points in which the normals drawn from
any point on a given straight line meet an ellipse is a pair of conies.
7. If the normals at points of an ellipse, whose eccentric angles
are a, j8, and y, meet in a point, prove that
sin (j8 + 7) + sin (7 + a) + sin (a + iS)=0.
Hence, by page 235, Ex. 15, shew that if PQB be a maximum
triangle inscribed in an ellipse, the normals at P, Q, and B are
concurrent.
[EXS. XLVI.] CONCURRENT NORMALS. EXAMPLES. 391
8. Prove that the normals at the points where the straight line
X y x^ v^
+ i~?- — = 1 meets the ellipse -^ + ^ =1 meet at the point
a cos a &sma ^ a^ b^
I - ae^ cos-* a, -J— sin** a 1
9. Prove that the loci of the point of intersection of normals at
the ends of focal chords of an ellipse are the two ellipses
aY ( 1 + e2)2 + 62 (a; ± ae) {x t ae^) = 0.
10. Tangents to the ellipse —2 + j^ = ^ are drawn from any point
on the ellipse -2 + f^=^; prove that the normals at the points of
contact meet on the ellipse a^x^ + bY=^{a^-b^)^.
11. . Any tangent to the rectangular hyperbola 4:xy=ab meets the
ellipse -g + |j=l in the points P and Q; prove that the normals at P
and Q meet on a fixed diameter.
12. Chords of an ellipse meet the major axis in the point whose
distance from the centre is a \ / ; prove that the normals at its
V a + b
ends meet on a circle.
13. From any point on the normal to the ellipse at the point
whose eccentric angle is a two other normals are drawn to it ; prove
that the locus of the point of intersection of the corresponding
tangents is the curve
xy + bxsina + ay cosa = 0.
14. Shew that the locus of the intersection of two perpendicular
normals to an ellipse is the curve
(a2 + 62) (a;2 + ^2) (^2^2 + 62^.2)2 _ (^2 _ ^2)2 (^2^2 _ 62^.2)2^
/i*2 nj2
15. ABC is a triangle inscribed in the elUpse —2 + 72=1 having
each side parallel to the tangent at the opposite angular point ; prove
that the normals at A, B, and G meet at a point which lies on the
eUipse a2a;2 + 62^2 = ^ (a2 - 62)2.
16. The normals at four points of an ellipse meet in a point {h, k).
Find the equations of the axes of the two parabolas which pass
through the four points. Prove that the angle between them is
2 tan-i - and that they are parallel to one or other of the equi-con-
jugates of the ellipse.
892 COORDINATE GEOMETRY. [EXS. XLVI.]
17. Prove that the centre of mean position of the four points on
le ellipse -o + |o =
(a, /3), is the point
the ellipse -2 + I2 — ^» *^® normals at which pass through the point
18. Prove that the product of the three normals drawn from any
point to a parabola, divided by the product of the two tangents from
the same point, is equal to one quarter of the latus rectum.
19. Prove that the conic 2aky = {2a-h)y^ + 4:ax^ intersects the
parabola y^=4:ax at the feet of the normals drawn to it from the point
{h, k).
20. From a point {h, k) four normals are drawn to the rectangular
hyperbola xy = c^; prove that the centre of mean position of their feet
(h k\
- , 2 ) J aiid that the four feet are such that each is the
orthocentre of the triangle formed by the other three.
Confocal Conies.
415. Def. Two conies are said to be confocal when
they have both foci common.
To find the equation to conies which are confocal with
the elli2^se
2 2
All conies having the same foci have the same centre
and axes.
The equation to any conic having the same centre and
axes as the given conic is
?4=i ••••■••;;;;;; (^)-
The foci of (1) are at the points {^\Ja^ — h^^ 0).
The foci of (2) are at the points {^sJA - B^ 0).
These foci are the same if
A-B = a^-h%
i.e.ii A-a' = B-b''^X (say).
.*. A^a^ + \, and B = ¥ + X.
CONFOCAL CONICS. 393
The equation (2) then becomes
^ 19 . \ -^5
which is therefore the required equation, the quantity A,
determining the particular confocal.
416. For different values of X to trace the conic given
hy the equation
^ + -^-=1 (1).
First, let X be very great ; then a^ + X and 6^ + X are
both very great and, the greater that X is, the more nearly
do these quantities approach to equality. A circle of
infinitely great radius is therefore a confocal of the
system.
Let X gradually decrease from infinity to zero ; the
semi-major axis \J a'^ + X gradually decreases from infinity
to a, and the semi-minor axis from infinity to h. When X
is positive, the equation (1) therefore represents an ellipse
gradually decreasing in size from an infinite circle to
the standard ellipse
a" b^
This latter ellipse is marked / in the figure.
Next, let X gradually decrease from to — b^. The
semi-major axis decreases from a to \/a^ — b^, and the semi-
minor axis from b to 0.
For these values of X the confocal is still an ellipse,
which always lies within the ellipse /; it gradually
decreases in size until, when X is a quantity very slightly
greater than — b^, it is an extremely narrow ellipse very
nearly coinciding with the line SH, which joins the two
foci of all curves of the system.
Next, let X be less than — b^ ; the semi-minor axis
\/b^ + X now becomes imaginary and the curve is a hyper-
bola ; when X is very slightly less than — b^ the curve is a
394
COOKDINATE GEOMETRY.
hyperbola very nearly coinciding with the straight lines
SX and SX'.
[As X passes through the value — h^ it will be noted that
the confocal instantaneously changes from the line-ellipse
SH to the line-hyperbola SX and HX'.^
As X gets less and less, the semi-transverse axis Ja^ + \
becomes less and less, so that the ends of the transverse
axis of the hyperbola gradually approach to C, and the
hyperbola widens out as in the figure.
When X = — a^, the transverse axis of the hyperbola
vanishes, and the hyperbola degenerates into the infinite
double line TOY'.
When X is less than — a% both semi-axes of the conic
become imaginary, and therefore the confocal becomes
wholly imaginary.
417. Through any point in the plane of a given conic
there can he drawn two conies confocal with it; also one of
these is an ellipse and the other a hyperbola.
Let the equation to the given conic be
t + t^l
and let the given point be (f g).
CONFOCAL CONICS. 395
Any conic confocal with the given conic is
If this go through the point (/, g), we have
•^ +^^ = 1 (2).
This is a quadratic equation to determine X and there-
fore gives two values of X.
Put b^ + X = fi, and hence
a^ + X = fi + a^-b^ = fjL + aV.
The equation (2) then becomes
i.e. />c2 + /x(aV-/2-/)-^2^iV = (3).
On applying the criterion of Art. 1 we at once see that
the roots of this equation are both real.
Also, since its last term is negative, the product of
these roots is negative, and therefore one value of ju, is
positive and the other is negative.
The two values of b^ + X are therefore one positive and
the other negative. Similarly, the two values of a^ + X can
be shewn to be both positive.
On substituting in (2) we thus obtain an ellipse and a
hyperbola.
418. Confocal conies cut at right angles.
Let the confocals be
+ 70 . V ^ 1, and -y— T-+TT-r^=l'
«2 + Xi b'^ + X^ ' a'^ + X^ b'^ + X,
and let them meet at the point ix\ y').
The equations to the tangents at this point are
xx' yy' - , £C£c' yy' ^
396 COORDINATE GEOMETRY.
These cut at right angles if (Art 69)
+ 7JJ-^v€T^^v^ = ^ «•
But, since (x\ 2/) is a common point of the two confocals,
we have
^ + ^ - 1 and 4- ^ - 1
«2 + ^^ + 52 + x,~'' ^'^"^ a' + X, h' + X,~ '
By subtraction, we have
/2/_i l_Vv'^^-^ ^-A=o
a;-
/2
**^' (a^ + X^ {a\-\- X,) ■*" (F+ Aj) (6^ + X,) " ^ ^*
The condition (1) is therefore satisfied and hence the
two confocals cut at right angles.
Cor. From equation (2) it is clear that the quantities
b^ + A-i and b^ + X^ have opposite signs ; for otherwise we
should, have the sum of two positive quantities equal to
zero. Two confocals, therefore, which intersect, are one an
ellipse and the other a hyperbola.
419. One conic and only one conic, confocal with the conic
-2 + 12 = I5 ^^^ ^^ drawn to touch a given straight line.
Let the equation to the given straight line be
X cos a + y sin a =p (1) .
Any confocal of the system is
x^ , y
+ ./:^ = l (2).
The straight line (1) touches (2) if
2j2= {a? + X) cos2 a + (&2 + X) sin2 a (Art. 264) ,
i.e. if X =_p2 _ a? cos^ a - 6^ sin2 ^^
This only gives one value for X and therefore there is only one
conic of the form (2) which touches the straight line (1).
Also X + a,2=^2^^^2_ j2^ gin2 jj_a j.gal quantity. The conic is
therefore real.
CONFOCAL CONICS. EXAMPLES. 397
EXAMPLES. XLVII
1. Prove that the difference of the squares of the perpendiculars
drawn from the centre upon parallel tangents to two given confocal
conies is constant.
2. Prove that the equation to the hyperbola drawn through the
point of the ellipse, whose eccentric angle is a, and which is confocal
with the elhpse, is
cos^ a sin^ a
3. Prove that the locus of the points lying on a system of confocal
eUipses, which have the same eccentric angle a, is a confocal hyperbola
whose asymptotes are inclined at an angle 2a.
4. Shew that the locus of the point of contact of tangents drawn
from a given point to a system of confocal conies is a cubic curve,
which passes through the given point and the foci.
If the given point be on the major axis, prove that the cubic
reduces to a circle.
5. Prove that the locus of the feet of the normals drawn from a
fixed point to a series of confocals is a cubic curve which passes
through the given point and the foci of the confocals.
6. A point P is taken on the conic whose equation is
such that the normal at it passes through a fixed point {h, k); prove
that P lies on the curve
1 — ? — = .
y -k x-h hy-kx
7. Two tangents at right angles to one another are drawn from
a point P, one to each of two confocal ellipses ; prove that P Hes on
a fixed circle. Shew also that the line joining the points of contact is
bisected by the line joining P to the common centre.
8. From a given point a pair of tangents is drawn to each of a
given system of confocals ; prove that the normals at the points of
contact meet on a straight Une.
9. Tangents are drawn to the parabola y^=4iXs/oP-h^, and on
each is taken the point at which it touches one of the confocals
a^ + \ b^+\
prove that the locus of such points is a straight line.
398 COORDINATE GEOMETRY. [Exs. XLVII.]
10. Normals are drawn from a given point to each of a system of
eonfocal conies, and tangents at the feet of these normals ; prove that
the locus of the middle points of the portions of these tangents
intercepted between the axes of the confocals is a straight line.
11. Prove that the locus of the pole of a given straight line with
respect to a series of confocals is a straight line which is the normal
to that eonfocal which the straight line touches.
12. A series of parallel tangents is drawn to a system of eonfocal
conies ; prove that the locus of the points of contact is a rectangular
hyperbola.
Shew also that the locus of the vertices of these rectangular
hyperbolas, for different directions of the tangents, is the curve
r'^ = c^eos2d, where 2c is the distance between the foci of the
confocals.
13. The locus of the pole of any tangent to a eonfocal with respect
to any circle, whose centre is one of the foci, is obtained and found to
be a circle ; prove that, if the circle corresponding to each eonfocal be
taken, they are all coaxal.
14. Prove that the two conies
ax^ + 2hxy + by^=l and a'x^ + 2h'xy + b'y^=l
can be placed so as to be eonfocal, if
(ah-h?f ~ {a'b'-h'Y '
Curvature.
420. Circle of Curvature. Def. If F, Q, and R
be any three points on a conic section, one circle and only
one circle can be drawn to pass through them. Also this
circle is completely determined by the three points.
Let now the points Q and R move up to, and ultimately
coincide with, the point P ; then the limiting position of
the above circle is called the circle of curvature at P ; also
the radius of this circle is called the radius of curvature at
Pj and its centre is called the centre of curvature at P.
421, Since the circle of curvature at P meets the
conic in three coincident points at P, it will cut the curve
in one other point P'. The line PP' which is the line
joining P to the other point of intersection of the conic and
the circle of curvature is called the common chord of
curvature.
I
CIRCLES OF CURVATURE. 399
We shewed, in Art. 400, that, if a circle and a conic
intersect in four points, the line joining one pair of points
of intersection and the line joining the other pair are
equally inclined to the axis. In our case, one pair of
points is two of the coincident points at P, and the line
joining them therefore the tangent at P ; the other pair of
points is the third point at P and the point P', and the
line joining them the chord of curvature PP'. Hence the
tangent at P and the chord of curvature PP' are, in any
conic, equally inclined to the axis.
4t^2i, To find the equation to the circle of curvature and
the length of the radius of curvature at any point (af, 2at)
of the parabola y"^ = 4:ax.
If S=0 be the equation to a conic, T=0 the equation
to the tangent at the point P, whose coordinates are at^ and
2at, and L = the equation to any straight line passing
through P, we know, by Art. 384, that jS + \.L.T=0 is
the equation to the conic section passing through three
coincident points at P and through the other point in which
X = meets aS^-0.
If A. and L be so chosen that this conic is a circle, it will
be the circle of curvature at P, and, by the last article, we
know that L = will be equally inclined to the axis with
T=0.
In the case of a parabola
S = 2/2 - iax, and T=ty-x- af. (Art. 229.)
Also the equation to a line through {af, 2at) equally
inclined with :Z' = to the axis is
t(y— 2at) + x — at'~0,
so that L ^ ty + X — Sat^.
The equation to the circle of curvature is therefore
y^ — 4ax ■{■ \(ty — X — af) {ty + x — ^af) = 0,
where 1 + Xi^ = — A., i.e. \ = —^ ^ •
I +f
400 COORDINATE GEOMETRY.
On substituting this value of A,, we have, as the required
equation,
x^ + y'- "lax {W + 2) + iayt^ - ZaH"" = 0,
i.e. [x-a{2 + 3f)J + [y + 2at^f = ia? (1 + ff.
The circle of curvature has therefore its centre at
the point (2a + ^af, — 2af) and its radius equal to
2a (1 + ff.
Cor. If S be the focus, we have SP equal to a + at\ so
2 . SP^
that the radius of curvature is equal to
V
a
423. To find the equation to the circle of curvature at
the 'point P (a cos <^, h sin ^) of the ellipse -3 + rg = 1-
a
The tangent at the point P is
X It
— cos <f> + T sin d) = 1.
a
The straight line passing through P and equally inclined
with this line to the axis is
cos<^. ,. sin<^ I ' ,\ n
(x — a cos ch) — ^ (y — o sm d>) = 0,
a
X 1/
i. e. - cos </> — r sin d> — cos 2d) = 0.
a
The equation to the circle of curvature is therefore of
the form
,2
^ y^
1 + X - cos ^ + f sin ^ — 1
- cos </) — ^ sin (^ - cos 2</) =0 (1).
Since it is a circle, the coefficients of x'^ and y^ must be
equal, so that
1 cos^ ^ 1 . sin^ <^
-g + A r— = 77; — A
and therefore X =
6^ cos^ cf> + a^ sin^ ^ *
CIRCLES OF CURVATURE. 401
On substitution in (1), the equation to the circle of
curvature is
+ («^ - b') [^ cos^ .f,-^ sin^ <l> - ^^ (1 + cos 2,^)
+ z (1 - cos 2cf>) + cos 2(j>\ = 0,
+ ^2 (cos^ <^ - 2 sin^ <^) - &2 (2 cos^ </, - sin^ </>) - 0.
The equation to the circle of curvature is then
^x -— cos^ ^1 + h/+ ~y- sm^ ^1
+ &^{2cos2<^-sin2^}
(a^ sin^ (i) + 6^ cos^ 6Y i, , .
= ^7^ , alter some reduction.
The centre is therefore the point whose coordinates are
cos^ ^, 7 — sin^ ^ j and whose radius is
(g^ sin^ <j> + ¥ cos^ <j!>) ^
Cor. 1. If Ci) be the semi-diameter which is conju-
gate to CP, then D is the point (90° + <^), so that its
coordinates are — a sin ^ and b cos <^. (Art. 285.)
Hence CD^ = a^ sin^ <^ + ¥ cos^ ^,
and therefore the radius of curvature p = — 5— .
ao
Cor. 2. If the point P have as coordinates x' and 3/'
then, since x' =^a cos <^ and 2/' = 5 sin <^, the equation to the
circle of curvature is
\^-^rn ^[y^-w-y) ^^-^•
L. 26
402 COORDINATE GEOMETRY.
Cor. 3. In a similar manner it may be shewn that the
equation to the circle of curvature at any point {x\ y') of
9 9
%)C II
the hyperbola — — — — 1 is
a
y ) - ^aW
424. If a circle and an ellipse intersect in four points,
the sum of their eccentric angles is equal to an even
multiple of tt. [Page 235, Ex. 18.]
If then the circle of curvature at a point P, whose
eccentric angle is 6, meet the curve again in Q^ whose
eccentric angle is ^, three of these four points coincide at
P, so that three of these eccentric angles are equal to 0,
whilst the fourth is equal to ^. We therefore have
3^ + <^ = an even multiple of tt =::: 2mr.
Hence, if <^ be supposed given, i.e. if Q be given, we
u ^ 2mr-cf>
have tf — o •
Giving n in succession the values 1, 2, and 3, we see
, , - T 27r - d> 4:7r-di Qtt - cj>
that 6 equals — ^— ^ , — ^ — , or — ^— .
Hence the circles of curvature at the points, whose
^ 27r — cf) 4:7r — d> , Gtt — </) „
eccentric angles are — - — , — ^ — » and — - — , all
pass through the point whose eccentric angle is cf).
Also since
27r — d) 47r — d) Gtt — d) , , ,,. , n
r + — —J- + — -— i- + ^ = 4r7r = an even multiple ot tt,
3 3 o
, - . , 27r— d) 4^7r — (b Gtt — d)
we see that the points — - — , — ^ — , — ^ — j and </>
all lie on a circle.
EVOLUTE OF A CURVE.
403
Hence through any point Q on an ellipse can be drawn
three circles which are the circles of curvature at three
points Pj, /*2> <^^^ -^3' -Also the four points Pj, Pg, P3, and
Q all lie on another circle.
425. E volute of a Curve. The locus of the
centres of curvature at different points of a curve is called
the evolute of the curve.
426. Evolute of the parabola 'if=A:ax.
Let (x, y) be the centre of curvature at the point (<x^^, 2a^)
of this curve.
Then ic = a (2 + Zf) and ^ = - "latK (Art. 422.)
.-. {x-2ay=21aH'=^^ay\
i.e. the locus of the centre of curvature is the curve
21ay^ = i{x~2ay.
This curve meets the axis of x in the point (20-, 0).
It also meets the parabola
where
27a2aj = (a;-2a)^
i. e. where x = ^a,
and therefore
y = ^4.J2a.
Hence it meets the parabola at
the points
(8a, ±4^2«).
The curve is called a semi-
cubical parabola and could be shewn
to be of the shape of the dotted curve in the figure.
427.
XT 11
Evolute of the ellipse — „ + ^„= 1.
^ a^ ¥
If {x, y) be the centre of curvature corresponding to the
point {ct, cos ^, b sin ^) of the ellipse, we have
a'-b^
x =
cos^ <fi and y = —
a'
-¥
sin^ <fi.
26—2
404
COORDINATE GEOMETRY.
Hence
{axf + (hyf = {a? - h'f {cos^ «/, + sin^ </>} = {a' - h'f.
Hence the locus of the point (x, y) is the curve
{axf-\-(hyf = {a?-h'f.
This curve could be shewn to
be of the shape shewn in the figure
where
CL = CL'
a^-¥
a
and CM^CM'^
a'
The equation to the evolute of
the hyperbola would be found to
be
{axf-{hyf=^{a? + h'f.
428. Contact of different orders. If two conies,
or curves, touch, i.e. have two coincident points in common
they are said to have contact of the first order. The
tangent to a conic therefore has contact of the first order
with it.
If two conies have three coincident points in common,
they are said to have contact of the second order. The
circle of curvature of a conic therefore has contact of the
second order with it.
If two conies have four coincident points in common,
they are said to have contact of the third order. No
conies, which are not coincident, can have more than four
coincident points ; for a conic is completely determined if
five points on it be given. Contact of the third order is
therefore all that two conies can have, and then they are
said to osculate one another.
Since a circle is completely determined when three
points on it are given we cannot, in general, obtain a circle
to have contact of a higher order than the second with a
given conic. The circle of curvature is therefore often called
the osculating circle.
CONTACT OF DIFFERENT ORDERS. 405
In general, one curve osculates another when it has the
highest possible order of contact with the second curve.
429. Equation to a conic osculating another conic.
If S—0 be the equation to a conic and T=0 the
tangent at any point of it, the conic S = \T^ passes through
four coincident points of S=0 at the point where T—0
touches it. (Art. 385, lY.)
Hence S= \T^ is the equation to the required osculating
conic.
Ex. The equation of any conic osculating the conic
ax^ + 2hxy + bif-2fy = (1)
at the origin is
ax^ + 2hxy + b7f-2fy + \y^=0 , (2).
For the tangent to (1) at the origin iay = 0.
If (2) be a parabola, we have h^ = a{b + 'K), so that its equation is
{ax + hy)^=2afy.
If (2) be a rectangular hyperbola, we have a + b + \ = 0, and the
equation to the osculating rectangular hyperbola is
a {x^ - 2/2) + 2hxy- 2fy = 0.
EXAMPLES. XLVIII.
1. If the normal at a point P of a parabola meet the directrix in
L, prove that the radius of curvature at P is equal to 2PL.
2. If ft and P2 be the radii of curvature at the ends of a focal
chord of the parabola, prove that
3. PQ is the common chord of the parabola and its centre of
curvature at P ; prove that the ordinate of Q is three times that of P,
and that the locus of the middle point of PQ is another parabola.
4. If p and p' be the radii of curvature at the ends, P and D, of
conjugate diameters of the ellipse, prove that
and that the locus of the middle point of the line joining the centres
of curvature at P and D is
{ax + hy)^ + (aa; - hy)^ = (a^ •
406 COORDINATE GEOMETRY. [Exs.
5. is the centre of curvature at any point of an ellipse, and Q
and R are the feet of the other normals drawn from ; prove that the
locus of the intersection of tangents at Q and B is -3+ -2 = 1» ^.nd
X y
that the line QR is a normal to the ellipse
x^ «2 a^j)^
a2^ &2-(a2_ft2)2-
6. If four normals be drawn to an ellipse from any point on the
evolute, prove that the locus of the centre of the rectangular hyperbola
through their feet is the curve
©*-(!)*-•
7. In general, prove that there are six points on an ellipse the
circles of curvature at which pass through a given point O, not on the
ellipse. If be on the ellipse, why is the number of circles of
curvature passing through it only four?
8. The circles of curvature at three points of an ellipse meet in a
point P on the curve. Prove that (1) the normals at these three
points meet on the normal drawn at the other end of the diameter
through P, and (2) the locus of these points of intersection for
different positions of P is the ellipse
4(a%2 + &2^2)^(^2_62)2,
9. Prove that the equation to the circle of curvature at any point
{x', y') of the rectangular hyperbola x^-y^ = a^ is
a^ (a;2 + y^)- ^xx'^ + iyy'^ + Sa^ (a;'2 + y'^) = 0.
10. Shew that the equation to the chord of curvature of the
rectangular hyperbola xy = c^ at the point "«" is ty + t^x = c{l + t'^),
and that the centre of curvature is the point
V^-2^' '^'
Prove also that the locus of the .pole of the chord of curvature is
the curve r^ = 2c^ sin 29.
11. PQ is the normal at any point of a rectangular hyperbola and
meets the curve again in Q ; the diameter through Q meets the curve
again in R ; shew that FR is the chord of curvature at P, and that
PQ is equal to the diameter of curvature at P.
12. Prove that the equation to the circle of curvature of the conic
ax^+'2hxy + by^=i2y at the origin is
a{x'^ + y^) = 2y.
13. If two confocal conies intersect, prove that the centre of
curvature of either curve at a point of intersection is the pole of the
tangent at that point with regard to the other curve.
XLVIIIJ ENVELOPES. 407
14. Shew that the equation to the parabola, having contact of the
third order with the rectangular hyperbola icy = c^ at the point
(-. i)-
is {x-yt^)^-4ct{x + yt^) + 8cH^ = 0.
Prove also that its directrix bisects, and is perpendicular to, the
radius vector of the hyperbola from the centre to the point of contact.
15. ^ Prove that the equation to the parabola, which passes through
the origin and has contact of the second order with the parabola
y'^—^ax at the point (at^^ 2ai), is
(4a; - %tyf + 4a«2 (3a; - "Ity) = 0.
16. Prove that the equation to the rectangular hyperbola, having
contact of the third order with the parabola y^ = 4cax at the point
{at^^ 2af), is
x^ - 2txy - 2/2 + 2aa; (2 + 3«2) - 2at^y + a^t^ = 0.
Prove also that the locus of the centres of these hyperbolas is an
equal parabola having the same axis and directrix as the original
parabola.
17. Through every point of a circle is drawn the rectangular
hyperbola of closest contact; prove that the centres of all these
hyperbolas lie on a concentric circle of twice its radius.
18. A rectangular hyperbola is drawn to have contact of the third
X^ 7/2
order with the ellipse -^ + ^2 = ^ 5 ^^^ i*^ equation and prove that the
locus of its centre is the curve
/a;2 + i/2y
"^ ^ "^ &2 '
Envelopes.
430. Consider any point P on a circle whose centre
is and whose radius is a. The straight line through P
at right angles to OP is a tangent to the circle at P.
Conversely, if through we draw any straight line OP oi
length a, and if through the end P we draw a straight
line perpendicular to OP^ this latter straight line touches,
or envelopes, a circle of radius a and centre 0, and this
circle is said to be the envelope of the straight lines drawn
in this manner.
Again, if S be the focus of a parabola, and PY be the
tangent at any point P of it meeting the tangent at the
408 COORDINATE GEOMETRY.
vertex in the point Y, then we know (Art. 211, 8) that
STP is a right angle. Conversely, if S be joined to any
point J' on a given line, and a straight line be drawn
through Y perpendicular to SY, this line, so drawn, always
touches, or envelopes, a parabola whose focus is S and such
that the given line is the tangent at its vertex.
431. Envelope. Def. The curve which is touched
by each of a series of lines, which are all drawn to satisfy
some given condition, is called the Envelope of these
lines.
As an example, consider the series of straight lines
which are drawn so that each of them cuts off from a pair
of fixed straight lines a triangle of constant area.
We know (Art. 330) that any tangent to a hyperbola
always cuts off a triangle of constant area from its asymp-
totes.
Conversely, we conclude that, if a variable straight line
cut off a constant area from two given straight lines, it
always touches a hyperbola whose asymptotes are the two
given straight lines, i. e. that its envelope is a hyperbola.
432. If the equatio7i to any curve involve a variable
parameter, in the first degree only, the curve always passes
through a fixed point or points.
For if X be the variable parameter, the equation to the
curve can be written in the form S + \S' — 0, and this
equation is always satisfied by the points which satisfy
S = and S' = 0, i.e. the curve always passes through the
point, or points, of intersection oi S-Q and S' = [compare
Art. 97].
433. Curve touched hy a variable straight line whose
equation involves, in the second degree, a variable parameter.
As an example, let us find the envelope of the straight
lines given by the equation
m^x — my + a = (1),
where m is a quantity which, by its variation, gives the
series of straight lines.
ENVELOPES. 409
If (1) pass through the fixed point (A, ^), we have
w^h — mk + a = (2).
This is an equation giving the values of m correspond-
ing to the straight lines of the series which pass through
the point (A, li). There can therefore be drawn two
straight lines from (A, ^) to touch the required envelope.
As (A, Aj) moves nearer and nearer to the required
envelope these two tangents approach more and more
nearly to coincidence, until, when (A, h) is taken on the
envelope, the two tangents coincide.
Conversely, if the two tangents given by (2) coincide,
the point (A, U) lies on the envelope.
Now the roots of (2) are equal if l? = 4:ah,
so that the locus of (A, k), i. e. the required envelope, is the
parabola y^ = 4:ax.
Hence, more simply, the envelope of the straight line (1 )
is the curve whose equation is obtained by writing down
the condition that the equation (1), considered as a quad-
ratic equation in m, may have equal roots.
By writing (1) in the form
a
y = mx H — ,
m
it is clear that it always touches the parabola y^ = ^ax.
In the next article we shall apply this method to the
general case.
434. To find the envelope of a straight line whose
equation involves^ in the second degree, a variable parameter.
The equation to the straight line is of the form
X^P-{-\Q + R = (1),
where X is a variable parameter and P, Q^ and M are
expressions of the first degree in x and y.
Equation (1) may be looked upon as an equation
giving the two values of X corresponding to any given
point T.
410 COORDINATE GEOMETRY.
Through this given point two straight lines to touch the
required envelope may therefore be drawn.
If the point T be taken on the required envelope, the
two tangents that can be drawn from it coalesce into the
one tangent at T to the envelope.
Conversely, if the two straight lines given by (1)
coincide, the resulting condition will give us the equation
to the envelope.
But the condition that (1) shall have equal roots is
Q^ = 4.PR (2).
This is therefore the equation to the required envelope.
Since P, Q, and R are all expressions of the first degree,
the equation (2) is, in general, an equation of the second
degree, and hence, in general, represents a conic section.
The envelope of any straight line, whose equation
contains an arbitrary parameter and square thereof, is
therefore always a conic.
435. The method of the previous article holds even if
P, Q, and R be not necessarily linear expressions. It
follows that the envelope of any family of curves, whose
equation contains a variable parameter X, in the second
degree, is found by writing down the condition that the
equation, considered as an equation in X, may have equal
roots.
436. Ex. 1. Find the envelope of the straight line which cuts off
from two given straight lines a triangle of constant area.
Let the given straight lines be taken as the axes of coordinates and
let them be inclined at an angle w.
The equation to a straight line cutting off intercepts / and g from
the axes is
rl=' <^>-
If the area of the triangle cut off be constant, we have
\f .g . sin w = const.,
i.e, fg=zconst.=K^ (2).
On substitution for g in (1), the equation to the straight line
becomes f'y -fIO + K^x=:0.
ENVELOPES. EXAMPLES. 411
By the last article, the envelope of this line, for different values of
/, is given by the equation
I.e. xy=-^.
The result is therefore a hyperbola whose asymptotes coincide with
the axes of coordinates.
Ex. 2. Find the envelope of the straight line ivhich is such that
the product of the perpendiculars draiun to it from two fixed points is
constant.
Take the middle point of the line joining the two fixed points as
the origin, the line joining them as the axis of x, and let the two
points be {d, 0) and {-d, 0).
Let the variable straight line have as equation
y=mx + c.
The condition then gives
md + c -md + c , , ^„
X = constant = ^%
jjl + m^ Jl + m^'
so that c^ - viH^ =A^{1+ m^) .
The equation to the variable straight line is then
y-'mx=c= fJ{A'^ + d^)m^ + A^.
Or, on squaring,
mP (a;2 -A^- d^) - 2mxy + (y^ - A^) = 0.
By Art. 435, the envelope of this is
[2xyf = 4 (a;2 - ^2 _ ^2) (^2 _ ^2) ^
This is an ellipse whose axes are the axes of coordinates and whose
foci are the two given points.
Ex. 3. Find the envelope of chords of an ellipse the tangents at the
end of ivhich intersect at right angles.
Let the ellipse be — „ + ^r, = 1.
a^ 0^
If the tangents intersect at right angles, their point of intersection
P must lie on the director circle, and hence its coordinates must be of
the form (c cos ^, c sin 6), where c = Ja'^+ h^.
The chord is then the polar of P with respect to the ellipse, and
hence its equation is
x . c cos 6 V .c sin d ^
412 COORDINATE GEOMETRY.
a
Let t^tan - . Then since
l-*an2- ^_^2 2(
1 + tan^ -
the equation to the line is
ex 1-t'^ cy 2t _^
The envelope of this is (Art. 434),
^2g2 ^2g2
^2 2/^ _i
Since -:; — r-, s — r^ = a^ - &2 this equation represents a conic
a^ + b^ a^ + b^
confocal with the given one.
Ex. 4. The normals at four points of an ellipse meet in a point ;
if the line joining one pair of these points pass through a fixed point,
prove that the line joining the other pair envelopes a parabola which
touches the axes.
Let the equation to the ellipse be
%*t=^ W'
and let the equation to the two pairs of lines through the points be
lx+m7j = l (2),
and lj^x + miy = l (3).
By Art. 412, Cor. (1), we then have
11-^= — 2 ^^^ *^%=~i^ W*
If the straight line (3) pass through the fixed point (/, g), we have
BO that, by (4), -X_^|_=l,
and therefore 1= — ^ -r^ .
ENVELOPES. EXAMPLES. 413
If this value of I be substituted in (2), it becomes
m^a%^y + m {a?gy - W-fx - a^ft^) - d^g = 0,
the envelope of which is
(a^gy - hjx - a%Y = - ^a?g . a^&s^ ,
i.e, {a^gy - V^fxf + 2.a%^ {bjx + a?gy) + a%'^=0 (5) .
This is a parabola since the terms of the second degree form a
perfect square. Also, putting in succession x and y equal to zero, we
get perfect squares, so that the parabola touches both axes.
437. To find the envelope of the straight line
lx + TYiy + n — ^ (1),
where the quantities I, m, and n are connected hy the
relation
aP + hm? + cn^ + Ifmn + "Ignl + Vilm = (2).
[Equation (1) contains two independent parameters —
and - , whilst (2) is an equation connecting them. We
n
could therefore solve (2) to give - in terms of — : on sub-
^ ' n n
stituting in (1) we should then have an equation containing
one independent parameter and its envelope could then be
found.
It is easier, however, to proceed as follows.]
Eliminating n between (1) and (2), we see that the
equation to the straight line may be written in the form
aV' + htn^ ■\-g(}x-\- myY — 2 {fm + gl) (Ix + my) + 2hlm = 0,
/ ^\^ I
i. e. (a — 2gx + cx^) ( — j + 2 {cxy — gy -fx + h) —
^{h-2fy + cf) = 0.
The envelope of this is, by Art. 435,
{cxy - gy -fx + hf = {a- 2gx + ex") (b - 2fy + cy\
i.e., on reduction,
x^ (be -D + y' (ca - g') + 2xy (fg - ch)
+ 2x{fh-bg)^-2y{gh-af) + ab-h'' = 0.
414 COORDINATE GEOMETRY.
The envelope is therefore a conic section.
Cor. The envelope is a parabola if
i. e. if c = 0, or if abc + %fgh — af^ — hg^ - ch^ ~ 0.
438. Ex. Find the envelope of all chords of the parabola y^ = 4aa;
lohich subtend a given angle a at the vertex.
Any straight line is
Za3 + m7/ + w = (1).
The lines joining the origin to its intersections with the parabola
are, (by Art. 122), ny^ = - 4:ax {Ix + my) ,
i.e. ny^ + 4:a'mxy + 4alx^=0.
If a be the angle between these lines, we have
2/J'ia^m^-4aln
tana=— ^^ j-z ,
n+4:al
i.e. 16a2Z2 - IQa^ cot^ am^+n^ + 8aln{l + 2 cot^ a) = 0.
With this condition the envelope of (1) is, by the last article,
a;2 ( - 16a2 cot2 a) + 2/2 [I6a2 _ (4^ + 8a cot2 a)2]
+ 2x . 16a2 cot2 a (4a + 8a Cot2 a) - 256a4 cot2 a= 0,
i.e. the ellipse
[a; - 4a (1 + 2 cot2 a)]2 + 4 cosec2 a . 2/2= 64 cot2 a . cosec2 a.
EXAMPLES. XLIX.
Find the envelope of the straight line - + ^=1 when
a p
1. aa + b^=c. 2. a + ^+s/aF+^=c.
b^ a^
^' ^2 + ^2=1-
Find the envelope of a straight hne which moves so that
4. the sum of the intercepts made by it on two given straight
lines is constant.
5. the sum of the squares of the perpendiculars drawn to it from
two given points is constant.
6. the difference of these squares is constant.
7. Find the envelope of the straight line whose equation is
ax cos 9 + by sin 6 = c^.
[EXS. XLIX.] ENVELOPES. EXAMPLES. 415
8. Circles are described touching each of two given straight lines ;
prove that the polars of a given point with respect to these circles all
touch a parabola.
9. From any point P on a parabola perpendiculars P3I and FN
are drawn to the axis and tangent at the vertex; prove that the
envelope of MN is another parabola.
10. Shew that the envelope of the chord which is common to the
parabola y^=:4:ax and its circle of curvature is the parabola
y^ + 12ax = 0.
11. Perpendiculars are drawn to the tangents to the parabola
y^=4:ax at the points where they meet the straight line x = b; prove
that they envelope another parabola having the same focus.
12. A variable tangent to a given parabola cuts a fixed tangent in
the point A ; prove that the envelope of the straight line through A
perpendicular to the variable tangent is another parabola.
13. Shew that the envelope of chords of a parabola the tangents
at the ends of which meet at a constant angle is, in general an ellipse.
14. A given parabola slides between two axes at right angles ;
prove that the envelope of its latus rectum is a fixed circle.
15. Prove that the envelope of chords of an ellipse which subtend
a right angle at its centre is a concentric circle.
16. If the lines joining any point P on an ellipse to the foci meet
the curve again in Q and R, prove that the envelope of the line QE is
the concentric and coaxal ellipse
x^ y^ fl + ey_
17. Prove that the envelope of chords of the rectangular hyperbola
xy=a^, which subtend a constant angle a at the point (a?', y') on the
curve, is the hyperbola
x^x'^ + yh^'^=2a^xy (1 + 2 cot^ a) - 4a^ cosec^ a.
18. Chords of a conic are drawn subtending a right angle at a
fixed point 0. Prove that their envelope is a conic whose focus is
and whose directrix is the polar of with respect to the original conic.
19. Shew that the envelope of the polars of a fixed point with
respect to a system of confocal conies, whose centre is 0, is a parabola
having GO as directrix.
20. A given straight line meets one of a system of confocal conies
in P and Q, and BS is the line joining the feet of the other two
normals drawn from the point of intersection of the normals at P and
Q ; prove that the envelope of RS is a parabola touching the axes.
416 COORDINATE GEOMETRY. [ExS. XLIX.]
21. ABGD is a rectangular sheet of paper, and it is folded over so
that G lies on the side AB ; prove that the envelope of the crease so
formed is a parabola, whose focus is the initial position of G.
22. A circle, whose centre is A, is traced on a sheet of paper and
any point B is taken on the paper. If the paper be folded so that the
circumference of the circle passes through B, prove that the envelope
of the crease so formed is a conic whose foci are A and B.
23. In the conic -=l-ecos^ find the envelope of chords which
r
subtend a constant angle 2a at the focus.
24. Circles are described on chords of the parabola y^ = 'iax, which
are parallel to the straight line lx + my = 0, as diameters; prove that
they envelope the parabola
[ly + 2maf = 4a (Z^ + m^) {x + a).
25. Prove that the envelope of the polar of any point on the circle
{x-\-af+{y + hf=h'^ with respect to the circle x^+y^=c^ is the conic
h^ (a;2 + 2/2) = [ax + hy + c^-
26. Chords of the conic -=l-ecos^ are drawn passing through
r
the origin and on these circles as diameters circles are described.
Shew that the envelope of these circles is the two circles
- — h ecoBd ) = l±e.
r \r J
ANSWERS.,
I. (Pages 14, 15.)
1. 5. 2. 13. 3. 3V7. 4. s/aJ+b\
5. ^a2 + 2&2 + c2-2afc-26c. 6. 2a sin -^^.
7. a {j)x^ - m^) sj{m^ + m.^f + 4. 9. 3±2^15.
15. (¥->¥)• 16. (-2,-9). 17. (1, -I); (-11, 16).
18. (-5H, 2A); (-20i, 34|). 19. (-i,0); (-1, 2).
20. (-1,1); (1.1); (I, -I).
^^- \'a + b' ^+& 7' V«-^ ' a-bj'
^^- \^ /c + Z + m ' h + l + m J'
II.' (Pages 18, 19.)
1. 10. 2. 1. 3. 29. 4. 2ac.
5. a?. 6. 2fl& sin ^IZ,^? gin ^^I^^ sin ^LlL^a .
i Jj a
7. a2(w2-»n3)(m3-mi)(mi-W2).
8. la^im^-m^im^-m-^ivi-^-m^.
9 . la'^ (mg - W3) (wig - wij) (m^ - mg) -f- mjin^m^ .
13. 201. 14, 96.
III. (Pages 22, 23.)
12. 2^5. 13. x/79. 14. s/7a. 16. 1(8-3^3).
17. '^-. 18. i«V3- 25. r^=ci\ 26. ^=a.
27. r=2acose. 28. r cos 2^ = 2a sin ^. 29. r cos ^ = 2a sin^ ^.
30. r2 = a2 cos 2^. 31. x^ + 2f = a^. 32. y = mx.
33. a;2 + 2/2 = aa;. 34. {x^ + 2ff = 4:a^xY-
35. (a;2 + 2/2)2z=a2(a;2-2/2). 35, a;i/ = a2. 37. x'^-if = a\
38. 2/2 + 4flx = 4a2. 39. 2/2 = 4aa; + 4a2.
40. ^^ - 3x^2 ^ 3a;22^ -y^= 5kxy.
L. 27
11 COORDINATE GEOMETRY.
IV. (Page 30.)
8. 2ax + k^-=^0. 9. Oi2-l)(a;2 + ?/+a2) + 2aa;{w2 + l) = 0.
10. 4a;2(c2-4a2) + 4cV = c2(c2-4a2). H. {Qa-2c) x = a^-c\
12. ^f-i^J-2x + 5 = 0. 13. ^y + 2x + S = 0. 14. x + y = l.
15. y = x. 16. ?/ = 3.r. 17. 15a;2-2/2 + 2aa; = a2.
18. .r2 + 2/2 = 3. 19. fl^2 + ^2^4y^
20. 8a:2 + 8?/2 + 6;r-36;c + 27=:0. 21. .'c2 = 3?/2.
22. X'-\-2ay = a'^.
23. (1) 4a;2+3?/2 + 2a2/ = a2; (2) x'' -^if + Say = 4.a^.
V. (Pages 41, 42.)
1. y=x + l. 2. a;-2/-5 = 0. 3. x-y >^^-2^2> = 0.
4. 52/-3a; + 9=:0. 5. 2a; + 3y = 6. 6. 6a; - 5?/ + 30 = 0.
7. (1) a; + 2/ = ll; (2) 2/-a;=l. 8. x^y + 1 = ; x-y = ^.
9. a;^' + a:V = 2.-cy. 10. 20?/ -9a; = 96. 15. x-\-y=0.
16. 2/-^ = l- 17. ly + 10x = ll.
18. ax - by = (lb. 19. («-2&) a;- &?/ + &^ + 2a&- a2 = 0.
20. ^j(«i + i2)-2a;=:2fl^if2. 21. «ii22/ + ^ = « {*i + ^2)-
22. X cos 1 (01 + 02) + 2/ sin i (01 + 02) = a cps J (01 - 02).
«« ^ 01 + 02 2/ • 01 + 02 01 ~ 02
23. -cos?^+|sin^^=cos'^-^--^.
24. &a; cos 4 (01 - 02) - «2/ sin J (0i + 02) = ab cos i (0i + 02).
25. x + Sy + 7 = 0; y-3x=l; y + 7x = ll.
26. 2x-Sy = 4:; y-Sx=l; x + 2y = 2.
27. 1/ (a' -a)-x {V -b) = a'b - ah' ; y {a' - a) J^x{b'-b) = a'b' - ab.
28. 2ay-2b'x = ab-a'b'. 29. y = ^x\ 2?/ = 3x.
VI. (Pages 48, 49.)
1. 90°. 2. tan-iff. 3. tan-i-f. 4. 60°.
8. 4y + 3a; = 18. 9. 7?/-8rc = 118. 10. 4i/ + lla;=10.
11. a; + 4?/ + 16 = 0. 12. a.r + ftr/ = a\
13. 2a; (a - a') + 2?/ (6 - &') = a^ - a"^ + 62- y^.
15. yx' -xy' = 0; a^xy' -b^x'y = {a^-b^)x'y' ; xx' -yy' = x'--y'^.
16. 121y-88a; = 371; 33?/ -24a; =1043.
17. a; = 3; 2/ = 4; 4i. 19. a; = ; 2/ + x/3a; = 0.
20. 2/ = ^; (l-m2)(?/-7f) = 2m(a;-/i).
21. tan-iH; 9a;-7i/ = l; 7a; + 9?/ = 73.
ANSWEES. m
VII. (Pages 53, 54.)
1. ^. 2. 2f. 3. 5^. 4.
5. a cos J (a -/S). 8.
a^ + ab-b^
Ja^ + b'-^
^1 + m^
9. |^(&±V^MT^), ol. 11. 4(2 + ^3).
VIII. (Pages 61-65.)
/ - 11 41\ _a&_ a& \
■'■• V 29" '297* a + 6 '« + &;•
,4. {a cos i (01 + 02) sec i (01 -0o), asini (01 + 0^)860 i(0i- 0^)}.
/ a{b-b') 2bb' \ 130
^" \ b + b' ' b + b'J' ^' 17^29'
8. y = a', %y = 4.x + ^a. 9. (1,1); 45°.
10. (f,i); tan-160. 11. (-1, -3); (3,1); (5,3).
12. (2, 1); tan-i,^. 13. 45°; (-5, 3) ; a;- 3^ = 9; 2.r-?/ = 8.
14. 3 and - f . 19. w?! («2 ~ ^3) + ^% (^3 ~ '^1) + "^3 (^1 ~ ^2) = 0-
20. (-4,-3). 21. (h/-!^)- ^ 23. 43.i;-29y = 71.
24. x-y=ii. 25. 2/ = 3jc. 26. y = x.
27. a^y-b^x=ab{a-b). 28. 3;r + 42/ = 5a. 29. a; + ?/ + 2 = 0.
30. 23a; + 23^ = 11. 31. 13.r - 23?/ = 64.
33. ^ic + 5j/ + + X (^'a; + J5 '?/ + C) = where A is
(1) -^,, (2) -;g7, (3) -^,--^, and (4) - ^.^.^^y^ ^..
37. y = 2', x = 6. 38. 99a; + 77?/ + 71 = 0; 7a;-9y-37=0.
39. x-2y + l = 0; 2x + y=:S.
40. ^(2V2-3)+?/(x/2-l) = 4V2-5;
a;(2V2 + 3) + ?/(V2 + l) = 4^2 + 5.
41. {y-b){m + m') + {x-a){l-mm') = 0;
[y - 6) (1 - ?u7?i') -{x- a) (m + ?/t') = 0.
42. 33a; + 9?/ = 31; 112ic- 64?/ + 141 = 0; 7y-x = 18.
43. a;(3+V17)+2/(5 + V17) = 15 + 4^17;
a;(4 + V10) + ?/(2 + ^10) = 4VlO + 12;
a; (2 V34 - 3 V5) +2/ ( V34 - 5V5) = 6 V34 - 5 V5.
44. A{y-k)-B{x-h)=^{Ax + By + C).
45. At an angle of 15° or 75° to the axis of x.
27—2
IV COORDINATE GEOMETRY.
IX. (Pages 72, 73.)
I. (1) tan-i^; (2) 15°. 2. tan"!^^.
3. tan-i^^^tancj.
7, y = 0, y = x-a, x = 2a, y = 2a, y = x + a, x = 0, y = x, x = a, and
y = a, where a is the length of a side.
10. 2/(6-V3) + ^(3V3-2) = 22-9V3. 11. h
12. 10?^ - 11a; + 1 = ; ^^ ^IH-
X. (Pages 78—80.)
4. (-7,3). 5. (-ii, If); m-
/-85 + 7V5 7V5-27 \ 35-7^5 „ n^.^.^.
^- \ 120"""' W^'J' 120" • '• ^^'S'^*'"^?-
(6 + ^/10 2 + VlO ] / 6-^10 2-V10\ /8-VlO lo + ^lO N
°- I 2 ' 2 j'V 2 ' 2"y'V 6 ' 6 J'
9. {%, f), (2, 12), (12, 2), and ( - 3, - 3) ; 1^2, 4^2, 4^2, and 6^2.
10. (-13^,194). 11. 4. 12. 7||. 13. f.
14. ^. 15. M&-c)(c-a)(a-&).
16. «^ (w^a - ^'^3) {>^h - ''^h) (^'h " 'm2)^2m^m^m^.
,« lf(C2-Cs)^ (Cs-Ci)2 (Ci-Co)2]
17. i(Ci-Co)2-f K-wi2- 18. oV-^ ^ + i-3 iL + U 2^1
23. (f,f)-
24. 10i/ + 32« + 43 = 0; 25a; + 29?/ + 5 = 0; 2/ = 5a; + 2; 52a; + 80i/ = 47.
26. (4 + iv/3, f + v/3) ; (4 + i^/3, f + W3).
XI. (Pages 85—87.)
1. a;2 + 2a;?/cota-2/2 = a2, 2. ?/2 + Xa;2=:Xa2.
3. (7?i + l)a;=(m-l)a. 4. (w + ?i) (a;2 + ?/'-' + a^)- 2aa;(m-/«) = c2.
5. .T + 2/ = csec2|. 6. a;-?/ = <icosec2-.
7, j; + ?/ = 2ccosecw. 8. 2/-^ = 2ccosecw,
9. a:2 + 2a;i/cosw + 2/^=4c2cosec2w.
10. (^^ + y^) cos w + a;?/ (1 + cos^ w) = a; [a cos w + &) + 2/ (& cos w + a).
II. a;(m + cosw)+i/(l+?«cosw) = 0.
12. (i) ai/2 + &a;2+(a + &)ic?/-a?/(a + 2&)-&a;(2rt + 6) + a&(a + 6) = 0;
(ii) y=a;. 19. A straight line.
20. A circle, centre 0. 25. A straight line.
27. If P he the point [h, k), the equation to the locus of S is
h k ,
- + - = 1.
a; 2/
ANSWERS. V
XII. (Page 94.)
1. {x-Sy){x-'iy) = 0;ta.n-'^^\. 2. {2.t- ll?/)(2a;-?/) = 0; tan-i -J.
3. (lla; + 2y)(3a;-7i/)=0; tan-iff. 4. x = l ; x = 2', x = 3.
5. 2/= ±4. 6. {y + ^x){y-2x){y-Sx) = 0;ia.n-^{-^);t&n-^{}).
7. a;(l-sin^)+?/cos^ = 0; a; {l + sin^) + ^cos^ = 0; 0.
8. r/siii^ + a;cos^= ±a;^/cos2^; tan-^ (cosec^ Aycos2^).
9. 12x^-7xy-12y^ = 0; Ux^ + Hxy -71y^ = 0\ x^-y"^0',
x^-y^=0.
XIII. (Pages 98, 99.)
1, (I, -¥);45«. 2. (2, 1); tan-if. 3. (-f, -t);90°.
4. (-1, 1); tan-13. 6. -15. 7. 2. 8. -10 or -171.
9. -12. 10. 6. 11. 6. 12. 14. 13. -3.
14. fori=f. 16. (i) c{a + h) = 0; {ii) e = 0, ox ae = bd.
17. Siz + e.'c^se; 5?/-6a;=14.
XV. (Page 112.)
1. (1) i/'2=4a;'; (2) 2x'^ + y"' = 6.
2. (1) a;'2 + 2/'2=:2ca;'; (2) x'^ + y'^=2cy\
3. (a-&)2(a;'2 + ^'2) = a262.
4. (1) 2x'y' + a^=0; 9x'^ + 25y'^ = 225', x'^ + y'^ = l.
5. x"^ + 2j"^ = r^', x'^-y'^=a^cos2a. 6. a;'^ - 4?/'2 = ^2.
8. tan-ij; -C-^^^M^a,
XVI. (Page 117.)
1. 2x'-^Qy' + l=::0. 2. a;'2 + ^3a;y = l. 3. x'^ + y'^-=8.
4. ?/'2=4a;'cosec2a.
XVII. (Pages 123—125.)
I x^ + y^ + 2x-4:y = L 2. a;2 + ?/2+10a; + 12?/ = 39.
3^ x^ + y^-2ax + 2hy = 2ab. 4. a;^ + 7/2 + 2fla; + 26i/ + 262 = 0.
5. (2,4); ^61. 6. (f, l);ix/l3. 7. (^ o) 5 ^ ^-
8. {o,-f);Jp+g'- 9. (-7--^. -7tT=0 ' '"
13. 15a;2 + 15?/2- 94.^ + 18^ + 55=0.
14. &(a;2 + 2/2-a2) = a;(&2 + ;i2_^2), ^5^ a:2 + 7/ - aa; - fcy = 0.
16. a;2 + 7/2-22a;-4?/ + 25 = 0. 17. x'^ + i/-5x-y + ^=0.
18. 3a;2 + 3?/2-29a;-19?/ + 56 = 0.
19. & (a;2 + 7/2) _ (a2 + &2) ^ + (a _ 6) (a2 + 1-) = 0.
21. a;2 + ?/2_3a;_42/=0.
VI COORDINATE GEOMETKY.
23. cc'^ + y^--hx-ky = 0. 24. x^ + if-2ijja^-b^=b'^.
25. a;2 + 2/2_iOa;-10?/ + 25 = 0. 26. a;2 + 2/2-2aa;-2a?/ + a2=0.
27. ^2 + 2/2 + 2(5^^12) {a; + 2/) + 37±10Vl2 = 0.
28. x^ + y^-ex + 4y + 9 = 0. 29. 6 (a;2 + 2/2)=a; (ft^ + c^).
30. a;2 + i/2+6V22/-6a; + 9 = 0.
31. x^ + y^-3x + 2 = 0; 2x^ + 2y^-5x- ^Sy + S = 0;
2x^ + 2y^-7x-jSy + Q = 0.
33. {a;+21)2 + (i/ + 13)2=652. 34. 8a;2 + 8y2 - 25a; -3^/ + 18=0.
36. a;2 + i/2=a2 + &2; x'^ + y^-2{a + b)x + 2{a-b)y + a'^ + h^ = 0.
XVin. (Pages 134, 135.)
1. 5a; -12^ = 152. 2. 24a; + 10?/ + 151 = 0.
3. a; + 2i/=±2V5. 4. x + 2y +g + 2f= ±J5 ^g^+f-c,
5. [-^^ ;j2)' ^- c = a;(0, &). 7. Yes.
8. fc = 40or-10. 9. a cos2 a + b sin2 a i s/a^ + &2 sin2 a.
10. Aa + Bb + C=:i=c >Ja^+B\
11. (1) y = mx±ajl + m^; (2) m?/ + a; = ± a ^/f+m^ ;
(3) ax±y Jb^-a?=ab', (4) a; + i/ = aV2.
12. 2^?-2-^2-p^2. 13. a;2 + ,/±V2aa; = 0; a;2 + 2/2±V2«?/ = 0.
14. c = 6-a7?i; c = b-am^ sj{l + m'^){a'^+b'^).
15. aj^ + ?/2-6a;-8?/+ff^ = 0.
16. a:2 + ?/2_2ca;-2c?/ + c2=0, where 2c=ia + b^ ^aFVb\
17. 5a;2 + 5?/2- 10a; + 30^ + 49 = 0. 18. a;2 + 2/2- 2ca;-2c?/ + c2=0.
19. {x-if-V{y-hY=r''. 20. x'' + y^-2ax-2^y = 0.
XIX. (Pages 144, 145.)
1. x + 2y = 7. 2. 8a;-2y = ll. 3. x = 0.
4. 23a; + 5?/ = 57. 5. by-ax = a\ 6. (5,10).
7. (I. -tU 8. (1,-2). 9. (i, -i).
10. (-2a, -26). 11. (6, --V-).
12. 3?/-2a;=13; (-W-,-W)- 13. (2,-1). 14. a;'2 + ^'23,2a2.
18. iV46. 19. 9. 20. V2a2 + 2a& + 62. 21. (¥, 2) ; |.
23. (1) 28a;2 + 33a;?/ -28?/2- 715a; -195^ + 4225 = 0;
(2) 123x2 - Uxy + 3?/2 _ C64a; + 226?/ + 763 = 0.
ANSWERS. Vll
XX. (Pages 147, 148.)
2. r- - 2ra cosec a . cos (^ - a) + a^ cot^ a = 0, r = 2a sin 6.
6. r2-r[acos(^-a) + &cos(^-jS)] + a6cos(a-j8) = 0.
8. &V + 2ac=:l.
XXI. (Page 149.)
1. 120O; (^-^^4^); ^-fV/^^^T^..
2. 30°; (8-6^3, 12-4^3); J^l -24.^%.
f g-fcoSQ} f-g cos o} \ ^ Jp + g^-2fgG0S(a
\ siu-^ w sm-^ w / sm w
4. x^- + ^2xy + y^--x{4: + 3^2)-2ij{3 + ^2) + S{J2-l)=0.
5. a;2 + a;?/ + 2/2 + 11a; + 13?/ + 18 = 0.
8. (a; - x') {x - x") + {y - y') (y - y") + cos w [{x - x') {y - y")
+ {x-x"){y-y')] = 0.
XXII. (Pages 156—159.)
4. A circle. 5. -A. circle. 6. -A. circle.
ft SlU 03
9. a;^ + 7/2 - 2a;?/ cos w= ^r — , the given radii being the axes.
11. A circle. 12. A circle.
16, (1) A circle ; (2) A circle ; (3) The polar of 0.
17. The curve ?' = a + acos^, the fixed point being the origin and
the centre of the circle on the initial line.
24. The same circle in each case.
33. 2ah^sJa^TV\ 35. a Vtt ; ^ = 4a; 63a;+16?/ + 100a = 0.
36. (i) x = Q, 3a; + 4i/ = 10, 2/ = 4, and 3?/ = 4a;.
(ii) 2/ = wia;+ c ^/l + ??^'^, where
±(& + c) ±(6-c)
XXIII. (Pages 164, 165.)
3. 3a;2 + 3?/2-8.v + 29?/ = 0. 4. lDX-lly = lU.
5. x + mj = 2. 6. 6x-7y + 12 = 0. 7. (-1,-1).
8. (If.ii^). 11. (A + l)(a;2 + 2/2) + 2\(a; + 2?/) = 4 + 6\.
13. {y-xf=0.
Vlll COORDINATE GEOMETRY.
XXIV. (Pages 172, 173.)
8. x^-y^ + 2mxy = c. 12. k{x^ + y'^) + {a-c)y-ck=0.
13. x^- + 7f-cx-by + a^=0. 14. x^ + y^-lQx-18y -4: = 0.
XXV. (Pages 178, 179.)
1. (7a; + 6?/)2- 570a; + 750?/ + 2100 = 0.
2. {ax- lyf - 2a^x - 2hhj + a^ + a%^ + 64=0.
3. (-1,2); y = 2', 4; (0,2). 4. (4, |) ; a; = 4; 2; (4,4).
5. («'|); ^ = «; 2a; (a, 0). 6. (1,2); 2/ = 2; 4; (0,2).
8. (i)i; (ii) 4. 9. (2,6). 11. y^-2x',y-12 = m{x-2i).
XXVI. (Pages 185—187.)
1. 4?/ = 3.'c + 12; 4a; + 3i/ = 34. 2. 4?/-a; = 24; 4a; + 2/=108.
3. 2/~^ = ^' y+x = ^', a; + 2/ + 3 = 0; a;-?/ = 9.
4. y = x; x + y = 4a; y + x = ; x-y = 4:a.
5. 4i/ = a; + 28; (28,14). 6. (|, ^) .
7. 2/ + 2^ + l = 0; (i, -2); 2z/ = a; + 8; (8,8).
8. (3a, 2^3a); ^|, -^aV 9. 4?/ == 9a; + 4 ; 4?/ = a; + 36.
13. (^^^«» aV275 + 2^; (3a, 2^3a).
14. h^y + a\v + aibi = 0. 15. a;=0.
XXVII. (Pages 197, 198.)
4. ix + Sy+l = 0. 5. 56t/ = 25.
XXVIII. (Pages 203—205.)
25. Take the general equation to the circle and introduce the
condition that the point (at'^, 2at) lies on it ; the sum of the
roots of the resulting equation in t is then found to be zero.
28. It can be shewn that the normals at the points "^i" and "ig"
meet on the parabola when tjt^=2 ; then use the previous
example.
XXIX. (Pages 209—211.)
1. y = bx. 2. cx = a. 3. y = ad.
4. y = [x-a) tan 2a. 5.2/^- ^^^ = 2ax.
6. x^ = iuL^i{x-af+y^]. 19. y^ = 2a{x-a).
^. 7. 7x^ + 2xy + 7y^ + 10x-10ij + 7 = 0. 8. Without.
ANSWERS. IX
20. y^-ky = 2a{x-h). 21. tf{tf-2ax + 4.a^-) + Sa'i^=0.
22. (8a2 + i/2_2aa;)2tan2a = 16a2(4aa;-?/).
23. y^ + 4:ay^{a-x)-lQa^x + an'^=0.
24. The parabola t/^ = 2a{x + 2a).
XXX. (Pages 214—216.)
1. y^ = a{x-a), 2. y^ = ^ax. 3. 27ay2 = (2.r ~ rt)(a;-5a)2.
4. A parabola. 5. A straight line.
6. 27at/^ -4 (a; -2aP=: constant. •
7. A straight line, itself a normal.
XXXII. (Pages 234, 235.)
1. (a) 3a;2 + 5?/=32; (/3) 3a;2 + 7?/ = 115.
2. 20a;2 + 36i/2 = 45. 3. x'^ + 2y'^=100. 4. 8.^2 + 9^/2= 1152.
5. (l)y; W6; (^iV6,0); (2)1; W^; (0,±tVV5);
(3) -V-; I; (0, 5) and (0, 1).
./3
9. x + 4V3t/ = 24V3; 11a; -4^3^ = 24^3 ; 7 and 13.
XXXIII. (Pages 245-248.)
1. x + ^y = 6', 9a;-3?/-5 = 0.
2. 25a; + 6?/ = 137; 6ic-25t/ + 20 = 0.
3. ±a;V7±4?/ = 16; ±4a;T2/V7-l V7.
5. ?/ = 3x±iVH^; (=^/^\/65, =F^VV195).
31. Use Arts. 145 and 260.
XXXIV. (Pages 262—264.)
1. x + 2y = L 2. 2a;-7'(/ + 8 = 0; (-1, -i).
3. 3a; + 8i/ = 9; 2a; = 3i/.
4. 9a;2- 24a;?/ -4^2 + 30a; + 402/ -55 = 0.
5. a2?/ + &2^ = 0; a2y-62a; = 0; ahj + h^x = 0\ ay + hx = 0.
XXXV. (Pages 268—270.)
1. x^ - 2xy cot 2a - if =a^-b'^. 2. cx^-2xy = ca^.
3. d2 (a;2 - a2)2 = 4 (62a;2 + a2|/2 - a262).
4. X (x2 - a2)2 ^ 2 (a;2?/2 + y^x^ + a22/2 - a2&2) .
X COORDINATE GEOMETKY.
5. (a;2+?/2_a2_52)2^4cot2a(&2.r2 + aV-«"&^)-
Q, ay = bxta,na. 7. &2a;2 + a2^2_4^2^2^
8. ¥x^ + aY=a^h^{a^ + h^). 9. h^x^ + aY=2a%y.
10. (62a:2 + a2l/2)2z=c2(6%2 + ^4y2).
11. (a2+&2) {b^x^ + aY)^ = a^^^{b^x^ + aY)-
12. fe^a; (a; - 7^) + ahj {y~k) = 0.
13. C2a262 (6%2 + a2j^2) + (^,2^2 _^ ^2^2 _ 1) (^)4a;2 4. ^4^2) ^ Q.
14. {b^x^ + ahjY = a^^H^^ + y^)'
15. a464 (a;2 + y2^ = (^2 + ^,2) (52^2 ^ ^2^2)2.
29. If the chords be PK and PK', let the equation to KK' be
y = mx + c ; transform the origin to P and, by means of Art. 122,
find the condition that the angle KPK' is a right angle ; substi-
tute for c in the equation to KK', and find the point of inter-
section of KK' and the normal at P. See also Art. 404.
XXXVI. (Pages 282—284.)
1. 16a;2-9i/2 = 36. 2. 25a;2- 144^2=900.
3. 65a;2-36?/2=441. 4. a;2-2/2 = 32.
5. 6, 4, {±^13, 0), 2|-. 6. 3a;2-2/2=3a2.
7. 7?/2 + 24.'c?/-24aa;-6a2/ + 15a2 = 0; (-|'«)5 12a; -9?/ + 29a =0.
8. (5, -V-)- 9. 242/-30a;=±V161.
14. 2/=±^±x/«"^^&^; {^'+^')\/^2^-
15. 9^ = 32;r. 16. 125a; -48^/ = 481.
29. (1) &%2+aV=«'&M&'-«'); (2) ^=«-5^2;
(3) a;2 (a2 + 2&2) _ a^ - 2«^^ea; + a^ {a^ - h^) = 0.
XXXVII. (Pages 295, 296.)
1. At the points (a, ±b>J2).
8. {2x + y + 2){x + 2y + l) = 0, (2a; + ?/ + 2) (.i; + 2?/ + l) = const.
9. 3a;2+ 10^;!/ + 8^/2 + 14x + 22i/ + 7 = 0;
3a;2 + lOcKy + 8i/2 + 14a: + 22i/ - 1 = 0.
XXXVIII. (Pages 302—305.)
16. (±fv/6a, =FtV6a); (^^W^^. ±x/6a).
XXXIX. (Pages 319—321.)
19. Transform the equation of the previous example to Cartesian
Coordinates.
ANSWERS. XI
XL. (Pages 331, 332.)
1. A hyperbola ; (2, 1) ; c'= - 26.
2. An ellipse; ( -|, -|); c'= -4. 3. A parabola.
4. A hyperbola ; ( - H, - -io) ; ^ = - 46.
5. Two straight lines ; (-H» if); ^' = 0.
6. A hyperbola; (-|i -gV); c'=~j\\.
7. {2x + By-l){4x-y + l) = 0; 8x^ + 10xij -Sy'^-2x + 'iy = 0.
8. {y + x-2){y-2x-S) = 0; y^-xy -2x^~5y + x + 18 = 0.
9. (lla;-2?/ + 4)(5a;-10y + 4) = 0;
55a;2 - 120x2/ + 20?/2 + 64a; - 48!/ + 32 = 0.
10. 19a;2 + 24a;?/ + 7/2-22.^-6^ + 4=0;
19a;2 + 24a;i/ + 7/2 _ 22a; - 67/ + 8 = 0.
12. a;2-7/2 = 4a2. 13. (aa;-67/)2 = (a2_^2) (^^ _ ;,a;).
14. {x-yf-2{x + y)+i = 0. 15. (.T3/ + a&)tan(a-(Q) = &.r-a7/.
16. ^2 + |^-2||cos(a-^) = sin2(a-/3). 17. A point.
18. Two straight lines. 19. A straight line and a parabola.
20. A straight line and a rectangular hyperbola.
21. A circle and a rectangular hyperbola.
22. A straight line and a circle.
23. Two imaginary straight lines.
24. A circle and a straight line. 25. A parabola.
26. A circle. 27. A hyperbola. 28. An ellipse.
XLI. (Pages 346—348.)
7 ( , ^^— ^ \ . 9. Two coincident straight lines.
'• V 676 ' 169/
10. tan^i=-|, tan^2 = fj »'i = \/3. and 7-3 =^*
11. ^1 = 45°, ^2 = 135°, ri=V2, and 7-2 = 2.
12. tan^, = 7 + 5V2; tan^2=7- 5;^2,
28. 2. 29. 5V\/3. 30. fV^-
31. (TyVVioTi, ^i^^VVio^); hJ^oT^io.
fa a ,„ 3rt , a ,_\ , ,„
32. (2-^4'^^' T'=2^^j' ^^^•
33. (-|=fW6, 4±iV6); K/3.
34. (--l±lv/6, l±tv/6); 2.
Xll COORDINATE GEOMETKY.
XLII. (Pages 354, 355.)
1. (1) 3; (2) 3; (3) 4; (4) 2; (5) 4; (6) 3; (7) 3.
10. Ax + Hy = and Hx + By = 0; II^ = AB, so that the conic is a
pair of parallel straight lines.
11. .r(.c + 3?/) = 0; (2a;-3?/)2=0.
XLIII. (Pages 363, 364.)
1. A conic touching 8 = where T=0 touches it and having its
asymptotes parallel to those of *S = 0.
A conic such that the two parallel straight lines u=Q and
?t + A; = pass through its intersections with 5 = 0.
XLIV. (Pages 375—377.)
6. (-1,5) and (4, -3). 7. (-|,-f). 8. (^^ ' ^) •
9. (-4, -4) and (-1, -1); x-\-y + l = and a; + i/ + 3 = 0.
15. If P be the given point, G the centre of the given director circle,
and PCP' a diameter, the focus S is such that PS.P'S is
constant.
16. If PP' he the given diameter and S a focus then PS.P'S is
constant.
XLV. (Pages 383, 384.)
5. Qx^ + 12xy + ly^-12x-lSy = 0.
17. The narrow ellipse (Art. 408), which is very nearly coincident with
the straight Hne BD, is one of the conies inscribed in the quadri-
lateral, and its centre is the middle point of BD. This middle
point, and similarly the middle points of -4C and OL, therefore
lie on the centre-locus.
XLVI. (Pages 390—392.)
7. Proceed as in Art. 413, and use, in addition, the second result
of Art. 412, Cor. 2. From the two results, thus obtained,
eliminate 8.
9. Take l^^x + vi^y -1 = (Art. 412, Cor. 1) as a focal chord of the
ellipse.
14. If the normals are perpendicular, so also are the tangents ; the
line IjX + nijy — 1 = is therefore the iDolar with respect to the
ellipse of a point {sja^ + b'^ cos,6, sja^ + b'-' sin.6) on the director
circle.
15. The triangle ABC is a maximum triangle (Page 235, Ex. 15)
inscribed in the ellipse.
20. Use the notation of Art. 333.
ANSWERS. Xlll
XL VII. (Pages 397, 398.)
11, The locus can be shewn to be a straight line which is perpendi-
cular to the given straight line ; also the given straight line
touches one of the confocals and its pole with respect to that
confocal is its point of contact ; this point of contact therefore
lies on the locus, which is therefore the normal.
14. As in Art. 366, use the Invariants of Art. 135.
XL VIII. (Pages 405—407.)
5. Two of the normals drawn from coincide, since it is a centre of
curvature. The straight line l■^x + 7n-j^^tJ = l (Art. 412) is therefore
a tangent to the ellipse at some point <p and hence, by Art. 412,
the equation to QB can be found in terms of (p.
XLIX. (Pages 414—416.)
c-
1. {by- ax- cy^='iacx. 2. x^ + y'^- c {x + y) + - = 0.
x^ w^
3. -^ + ^2=1. 4. A parabola touching each of the two lines.
5. A central conic. 6. A parabola. 7. a^x^ + h-y^ = c'^.
19. The line joining the foci is a particular case of the confocals and
the polar of O with respect to it is the major axis ; the minor
axis is another particular case, so that two of the polars are lines
through G at right angles ; also the tangents at to the con-
focals through it are two of the polars, and these are at right
angles. Thus both G and are on the directrix.
21. The crease is clearly the line bisecting at right angles the line
joining the initial position of G to the position which C occupies
when the paper is folded.
__ Zcosa ^ -
23. =1 -e cosa cos^.
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